leetcode动态规划(python与c++)
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leetcode动态规划(python与c++)
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1 .?斐波那契數(shù)??
class Solution:def fib(self, n: int) -> int:# if n==0:# return 0# elif n==1:# return 1# else:# return self.fib(n-1)+self.fib(n-2)a =0b =1for i in range(n):a,b = b,a+breturn a class Solution { public:int fib(int n) {int a = 0, b = 1;for(int i = 0; i < n; i++){int temp = a;a = a + b;b = temp;}return a;} };2. 第 N 個(gè)泰波那契數(shù)
class Solution:def tribonacci(self, n: int) -> int:if n == 0:return 0if n <= 2:return 1dp = [0]*(n+1)dp[1] = 1dp[2] = 1for i in range(3, n+1):dp[i] = dp[i-3] + dp[i-2] + dp[i-1]return dp[-1] class Solution { public:int tribonacci(int n) {if(n == 0){return 0;}if(n <= 2){return 1;}vector<int> dp(n + 1, 0);dp[1] = 1;dp[2] = 1;for(int i = 3; i < n+1; i++){dp[i] = dp[i-3] + dp[i-2] + dp[i-1];}return dp[n];} };3. 爬樓梯
class Solution:def climbStairs(self, n: int) -> int:if n < 3:return na = 1b = 2for i in range(3, n + 1):a, b = b, a + breturn b class Solution { public:int climbStairs(int n) {if(n < 3){return n;}int a = 1, b = 2;for(int i = 3; i < n + 1; i++){int temp = b;b = a + b;a = temp;}return b;} };?4. 使用最小花費(fèi)爬樓梯
class Solution:def minCostClimbingStairs(self, cost: List[int]) -> int:dp = [0]*len(cost)dp[0],dp[1] =cost[0],cost[1]for i in range(2,len(cost)):dp[i] = min(dp[i-1],dp[i-2])+cost[i]return min(dp[-1],dp[-2]) class Solution { public:int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();vector<int> dp(n, 0);dp[0] = cost[0];dp[1] = cost[1];for(int i = 2; i < n; i++){dp[i] = min(dp[i-1], dp[i-2]) + cost[i];}return min(dp[n-2], dp[n-1]);} };?5. 打家劫舍
class Solution:def rob(self, nums: List[int]) -> int:n = len(nums)if n < 2:return max(nums)dp = [0] * ndp[0] = nums[0]dp[1] = max(nums[1],nums[0])for i in range(2, n):dp[i] = max(dp[i-2]+nums[i], dp[i-1])return dp[-1] class Solution { public:int rob(vector<int>& nums) {int n = nums.size();if(n == 0){return 0;}if(n == 1){return nums[0];}if(n == 2){return max(nums[0],nums[1]);}vector<int> dp(n, 0);dp[0] = nums[0];dp[1] = max(nums[0], nums[1]);for(int i = 2; i < n; i++){dp[i] = max(dp[i-2]+nums[i], dp[i-1]);}return dp[n-1];} };6. 打家劫舍 II
class Solution:def rob(self, nums: List[int]) -> int:n = len(nums)if n < 2:return max(nums)dp1 = [0]*ndp2 = [0]*ndp1[0] = 0dp1[1] = nums[1]dp2[0] = nums[0]dp2[1] = max(nums[0], nums[1])for i in range(2, n):dp1[i] = max(dp1[i-2]+nums[i], dp1[i-1])for i in range(2, n-1):dp2[i] = max(dp2[i-2]+nums[i], dp2[i-1])return max(dp1[-1], dp2[-2])7. 刪除并獲得點(diǎn)數(shù)
class Solution:def deleteAndEarn(self, nums: List[int]) -> int:maxVal = max(nums)total = [0] * (maxVal + 1)for val in nums:total[val] += val# print(total)n = len(total)dp = [0]*nif n <=2:return max(total)dp[0] = total[0]dp[1] = max(total[0], total[1])for i in range(2, n):dp[i] = max(dp[i-1], dp[i-2] + total[i])return dp[-1]?8.?跳躍游戲
class Solution:def canJump(self, nums: List[int]) -> bool:#貪心算法most_dis = 0for i in range(len(nums)):if i <= most_dis:most_dis = max(most_dis, nums[i] + i)if most_dis >= len(nums) - 1:return Truereturn False class Solution { public:bool canJump(vector<int>& nums) {int most_length = 0;for(int i = 0; i < nums.size(); i++){if(i <= most_length){most_length = max(nums[i] + i, most_length);}if(most_length >= nums.size() - 1){return true;}}return false;int n = nums.size();vector<bool> opt(n, false);opt[0] = true;for(int i = 1; i < n; i++){opt[i] = opt[i - 1] && nums[i-1] >= 1;nums[i] = max(nums[i - 1] - 1, nums[i]);}return opt[n-1];} };9.?跳躍游戲 II
class Solution { public:int jump(vector<int>& nums) {int res = 0;int start = 0;int end = 1;int maxPos = 0;while(end < nums.size()){for(int i = start; i < end; i++){//能跳到的最遠(yuǎn)距離maxPos = max(maxPos, nums[i] + i);}start = end;//下一次起跳點(diǎn)范圍開(kāi)始的格子end = maxPos + 1;//下一次起跳點(diǎn)范圍結(jié)束的格子res++;}return res;} };10.?最大子序和
class Solution:def maxSubArray(self, nums: List[int]) -> int:for i in range(1,len(nums)):nums[i]+=max(nums[i-1],0)return max(nums) class Solution { public:int maxSubArray(vector<int>& nums) {int n = nums.size();int max_num = nums[0];for(int i = 1; i < n; i++){nums[i] += max(nums[i-1], 0);max_num = max(nums[i], max_num);}return max_num;} };11.?環(huán)形子數(shù)組的最大和
題解：
class Solution:def maxSubarraySumCircular(self, nums: List[int]) -> int:dpmax = nums.copy()dpmin = nums.copy()sum_ = sum(nums)for i in range(1, len(nums)):dpmax[i] = max(dpmax[i-1]+nums[i], dpmax[i])for i in range(1, len(nums)):dpmin[i] = min(dpmin[i-1]+nums[i], dpmin[i])# print(dpmax)# print(dpmin)max_value = max(dpmax)min_value = min(dpmin)if (sum_ - min_value) == 0:return max_valueelse:return max(max_value, sum_ - min_value) class Solution { public:int maxSubarraySumCircular(vector<int>& nums) {int sum_ = nums[0];vector<int>dpmax(nums);vector<int>dpmin(nums);for(int i=1;i<nums.size();i++){dpmax[i]=max(dpmax[i-1]+nums[i],nums[i]);dpmin[i]=min(dpmin[i-1]+nums[i],nums[i]);sum_ += nums[i];}int maxv=*max_element(dpmax.begin(),dpmax.end());int minv=*min_element(dpmin.begin(),dpmin.end());if(sum_ - minv == 0 ){return maxv;}else{return max(maxv, sum_ - minv);}} };12.?最佳觀光組合
class Solution:def maxScoreSightseeingPair(self, values: List[int]) -> int:# n = len(values)# max_score = float('-inf')# for i in range(n):# for j in range(i+1, n):# max_score = max(max_score, values[i]+values[j]+i-j)# return max_scoren = len(values)mx = values[0]max_score = float('-inf')for i in range(1, n):max_score = max(max_score, mx + values[i] - i)mx = max(mx, values[i] + i)return max_score class Solution { public:int maxScoreSightseeingPair(vector<int>& values) {int n = values.size();int mx = values[0];int max_score = INT_MIN;for(int i = 1; i < n; i++){max_score = max(max_score, mx + values[i] - i);mx = max(mx, values[i] + i);}return max_score;} };13.?買賣股票的最佳時(shí)機(jī)
class Solution:def maxProfit(self, prices: List[int]) -> int:if len(prices) <= 1:return 0min_price = prices[0]max_profit = 0for i in range(1, len(prices)):min_price = min(prices[i], min_price)max_profit = max(max_profit, prices[i] - min_price)return max_profit class Solution { public:int maxProfit(vector<int>& prices) {if(prices.size() <= 1){return 0;}int max_profit = 0, min_price = prices[0]; for(int i = 1; i < prices.size(); i++){min_price = min(min_price, prices[i]);max_profit = max(max_profit, prices[i] - min_price);}return max_profit; } };?14.?買賣股票的最佳時(shí)機(jī) II
class Solution:def maxProfit(self, prices: List[int]) -> int:if len(prices)<=1:return 0res = 0for i in range(1, len(prices)):res += max(0, prices[i] - prices[i-1])return res class Solution { public:int maxProfit(vector<int>& prices) {int n = prices.size();if(n <= 1){return 0;}int res = 0;for(int i = 1; i < n; i++){res += max(0, prices[i] - prices[i-1]);}return res;} };15.買賣股票的最佳時(shí)機(jī)含手續(xù)費(fèi)
class Solution:def maxProfit(self, prices: List[int], fee: int) -> int:dp = [[0 for i in range(2)] for i in range(len(prices))]dp[0][1] = -prices[0]for i in range(1, len(prices)):dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + prices[i] - fee)dp[i][1] = max(dp[i-1][1], dp[i-1][0]-prices[i])# print('==dp:', dp)return dp[-1][0]16.最佳買賣股票時(shí)機(jī)含冷凍期
?
class Solution:def maxProfit(self, prices: List[int]) -> int: if len(prices)<=1:return 0 dp = [[0 for i in range(2)] for i in range(len(prices))]# print('==dp:', dp)dp[0][1] = -prices[0]dp[1][0] = max(dp[1 - 1][0], dp[1 - 1][1] + prices[1])dp[1][1] = max(dp[1-1][1], -prices[1])for i in range(2, len(prices)):dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + prices[i])dp[i][1] = max(dp[i - 1][1], dp[i - 2][0] - prices[i])# print('==dp:', dp)return dp[-1][0]#返回沒(méi)有的最大利潤(rùn)16-1. 單詞拆分
思路１:動(dòng)態(tài)規(guī)劃
#動(dòng)態(tài)規(guī)劃 dp[i]表示 s 的前 i 位是否可以用 wordDict 中的單詞表示, # class Solution:def wordBreak(self, s, wordDict):n = len(s)dp = [False] * (n + 1)dp[0] = Truefor i in range(n):for j in range(i+1, n+1):if dp[i] and (s[i:j] in wordDict):dp[j] = Trueprint('==dp:', dp)return dp[-1] s = "leetcode" wordDict = ["leet", "code"] sol = Solution() res= sol.wordBreak(s, wordDict) print('==res:', res)c++實(shí)現(xiàn):
class Solution { public:bool wordBreak(string s, vector<string>& wordDict) {int n = s.size();unordered_set<string> wordDictSet;for (auto word: wordDict) {wordDictSet.insert(word);} vector<bool> dp(n+1, false);dp[0] = true;for(int i = 0; i < n; i++){for(int j = i+1; j < n+1; j++){ if(dp[i] && wordDictSet.find(s.substr(i, j - i)) != wordDictSet.end()) {// cout<<"s.substr(i, j - i):"<<s.substr(i, j - i)<<endl;dp[j] = true;}}}return dp[n];} };思路2：回溯加緩存
#遞歸 lru_cache用于緩存 將數(shù)據(jù)緩存下來(lái) 加快后續(xù)的數(shù)據(jù)獲取 相同參數(shù)調(diào)用時(shí)直接返回上一次的結(jié)果 import functools class Solution:@functools.lru_cache()def helper(self, s):if len(s) == 0:return Trueres = Falsefor i in range(1, len(s)+1):if s[:i] in self.wordDict:res = self.helper(s[i:]) or resreturn resdef wordBreak(self, s, wordDict):self.wordDict = wordDictreturn self.helper(s) s = "leetcode" wordDict = ["leet", "code"] # s = "aaaaaaa" # wordDict = ["aaaa", "aaa"] # s= "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab" # wordDict = ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"] sol = Solution() res= sol.wordBreak(s, wordDict) print('==res:', res)16-2.單詞拆分 II
思路：遞歸?
class Solution:def helper(self, s, wordDict, memo):if s in memo:#遞歸終止條件return memo[s]if s=='':#遞歸終止條件return []res = []for word in wordDict:if not s.startswith(word):continueif len(word)==len(s):#匹配上剛好相等res.append(word)else:#匹配上 但是字符還沒(méi)到最后rest = self.helper(s[len(word):], wordDict, memo)for tmp in rest:tmp = word+ " "+ tmpres.append(tmp)print('==res:', res)print('==memo:', memo)memo[s] = resreturn resdef wordBreak(self, s, wordDict):if s=='':return []return self.helper(s, wordDict, memo={}) s = "catsanddog" wordDict = ["and", "cat", "cats", "sand", "dog"] # s = "cat" # wordDict = ["cat"] sol = Solution() res = sol.wordBreak(s, wordDict) print(res)?17-1,盛最多水的容器
求Max{(j-i) * Min( h(i), h(j) )},?
height=[2,8,1,5,9,3,4]
暴力法： 超出時(shí)間限制
#解法1 class Solution:def maxarea(self,height):max_area=0for i in range(len(height)-1):for j in range(i+1,len(height)):if (j-i)*min(height[i],height[j])>max_area:max_area=(j-i)*min(height[i],height[j])index_i=iindex_j=jreturn index_i,index_j,max_areas=Solution() height=[2,8,1,5,9,3,4] i,j,max_area=s.maxarea(height) print(i,j,max_area)分析:暴力法時(shí)間復(fù)雜度為O(n2),想想看，
如果 h(7) >= h(1)，我們還有必要再遍歷h(6)，h(5)，...，h(2)嗎，其實(shí)不用，這便是暴力算法的冗余之處，多做了很多次無(wú)用的遍歷，i = 1這趟遍歷中，最大面積一定為 (7-1) * h(1) ；
如果 h(7) < h(1)，我們?cè)賴L試h(6)，如果h(6)>=h(1)，那么在i = 1這趟遍歷中的面積最大為(6-1) * h(1)，沒(méi)必要再試h(5)了，依次這樣下去。
動(dòng)態(tài)規(guī)劃:
面積最大值初始值設(shè)定 maxarea；
i, j 分別指向索引兩頭，動(dòng)態(tài)交替地調(diào)整 i, j ，進(jìn)而嘗試取得較大的相對(duì)高度，這個(gè)調(diào)整的策略是關(guān)鍵，同時(shí)，更新目標(biāo)函數(shù)即面積的最大值，如果大于maxarea，則更新；
直到 i >?j 為止；
返回最大值法
時(shí)間復(fù)雜度為 O(n)，空間復(fù)雜度為 O(1) 。?
#解法2 class Solution:def maxarea(self,height):left=0right=len(height)-1max_area=0while left<right:max_area = max(max_area,(right - left) * min(height[left], height[right]))if height[left]<height[right]:left+=1else:right-=1# index_i = left# index_j=rightreturn max_areas=Solution() height=[2,8,1,5,9,3,4] max_area=s.maxarea(height) print(max_area)17-2:接雨水
思路1.暴力法 對(duì)于i處能存的水，向右向左分別找到最大的值，在取這兩值中的最小值減去此刻的值就是能存的水,超時(shí)O(n^2)
class Solution:def trap(self, height):res = 0n = len(height)for i in range(1, n):print('==i:', i)left_max, right_max = 0, 0for j in range(i, -1, -1):#往左搜索left_max = max(left_max, height[j])for j in range(i, n):#往右搜索right_max = max(right_max, height[j])print('==left_max:', left_max)print('==right_max:', right_max)res +=min(right_max, left_max) - height[i]print('res:', res)return resheight = [0,1,0,2,1,0,1,3,2,1,2,1] sol = Solution() sol.trap(height)思路２.優(yōu)化，雙指針
#某個(gè)位置i處，它能存的水，取決于它左右兩邊(left_max,right_max)的最大值中較小的一個(gè)。 #對(duì)于位置left而言，它左邊最大值一定是left_max，右邊最大值“大于等于”right_max， # 這時(shí)候，如果left_max<right_max成立，那么它就知道自己能存多少水了。 # 無(wú)論右邊將來(lái)會(huì)不會(huì)出現(xiàn)更大的right_max，都不影響這個(gè)結(jié)果。 # 所以當(dāng)left_max<right_max時(shí)，我們就希望去處理left下標(biāo)，反之，我們希望去處理right下標(biāo)。O(n) class Solution:def trap(self, height):left,right =0,len(height)-1left_max,right_max =0,0res = 0while left<=right:if left_max <right_max:res+=max(0, left_max - height[left])left_max = max(left_max, height[left])left+=1else:res += max(0, right_max - height[right])right_max = max(right_max, height[right])right -= 1print('==res:', res)return resheight = [0,1,0,2,1,0,1,3,2,1,2,1] sol = Solution() sol.trap(height)c++實(shí)現(xiàn):
class Solution { public:int trap(vector<int>& height) {int left = 0, right = height.size() - 1;int left_max = 0, right_max = 0;int res = 0;while(left <= right){if(left_max < right_max){res += max(0, left_max - height[left]);left_max = max(height[left], left_max);left++;}else{res += max(0, right_max - height[right]);right_max = max(height[right], right_max);right--;}}return res;} };思路3:動(dòng)態(tài)規(guī)劃
開(kāi)出兩個(gè)數(shù)組，一個(gè)用于統(tǒng)計(jì)坐邊最大值，一個(gè)用于統(tǒng)計(jì)右邊最大值，這樣最終該點(diǎn)的雨水就是當(dāng)前點(diǎn)的短板減去當(dāng)前值。
class Solution:def trap(self, height: List[int]) -> int:length = len(height)if length == 0:return 0left_max = [0 for i in range(length)]left_max[0] = height[0]right_max = [0 for i in range(length)]right_max[-1] = height[-1]for i in range(1, length):left_max[i] = max(left_max[i-1], height[i])for i in range(length-2, -1, -1):right_max[i] = max(right_max[i + 1], height[i])res = 0for i in range(length):res += min(left_max[i], right_max[i]) - height[i]return resc++實(shí)現(xiàn) :
class Solution { public:int trap(vector<int>& height) {int res = 0;int length = height.size();if (length == 0){return res;}vector<int> left_max(length, 0);vector<int> right_max(length, 0);left_max[0] = height[0];right_max[length-1] = height[length-1];for (int i=1; i<length; i++){left_max[i] = max(left_max[i-1], height[i]);}for (int i=length-2; i>=0; i--){right_max[i] = max(right_max[i+1], height[i]);}for (int i=0; i<length; i++){res += min(right_max[i], left_max[i]) - height[i];}return res;} };18.?等差數(shù)列劃分
class Solution:def numberOfArithmeticSlices(self, nums: List[int]) -> int:n = len(nums)if n <= 1:return 0dis = nums[1] - nums[0]res = 0temp = 0for i in range(2, n):if nums[i] - nums[i-1] == dis:temp += 1else:dis = nums[i] - nums[i-1]temp = 0res += tempreturn res19.?解碼方法
class Solution:def numDecodings(self, s: str) -> int:n = len(s)f = [1] + [0] * nfor i in range(1, n + 1):if s[i-1] != '0':f[i] += f[i-1]if i > 1 and s[i-2] != '0' and int(s[i-2:i])<=26:f[i] += f[i-2]return f[n]20-1:丑數(shù)
思路:判斷是否能整除2,3,5依次整除下去，將不能整除和１進(jìn)行判斷就知道是否是丑數(shù)了
class Solution:def isUgly(self, n: int) -> bool:if n <= 0: return Falsewhile((n % 2) == 0):n /= 2while((n % 3) == 0):n /= 3while((n % 5) == 0):n /= 5return n == 1c++循環(huán)實(shí)現(xiàn):
class Solution { public:bool isUgly(int n) {if(n <= 0){return false;}while((n % 2) == 0){n /= 2;}while((n % 3) == 0){n /= 3;}while((n % 5) == 0){n /= 5;}return n == 1;} };c++遞歸實(shí)現(xiàn):
//遞歸寫法 class Solution { public:bool isUgly(int n) {if(n <= 0){return false;}while((n % 2) == 0){return isUgly(n / 2);}while((n % 3) == 0){return isUgly(n / 3);}while((n % 5) == 0){return isUgly(n / 5);}return n == 1;} };20-2.丑數(shù)
思路：題目要求的這個(gè)數(shù)字一定是由單個(gè)或者多個(gè)2,3,5的乘積，如果從小到大去枚舉在判斷是否由2,3,5乘積組成，工作量會(huì)很大，所以考慮用2,3,5從下往上遞推，需要開(kāi)辟空間為n的數(shù)組，采用動(dòng)態(tài)規(guī)劃,2,3,5分別有三個(gè)索引，如果滿足要求的數(shù)字等于2,3,5的倍數(shù)乘積，那么就直接將索引加1.
python代碼:
class Solution:def nthUglyNumber(self, n):dp, index_two, index_three, index_five = [1] * n, 0, 0, 0for i in range(1, n):two = dp[index_two] * 2three = dp[index_three] * 3five = dp[index_five] * 5dp[i] = min(two, three, five)if two==dp[i]:index_two+=1if three==dp[i]:index_three+=1if five==dp[i]:index_five+=1print('==dp:', dp)return dp[-1]n = 11 sol = Solution() sol.nthUglyNumber(n)c++代碼:
class Solution { public:int nthUglyNumber(int n) {vector<int> dp(n,1);int index_two=0;int index_three=0;int index_five=0;for (int i=1;i<n;i++){int two = dp[index_two]*2;int three = dp[index_three]*3;int five = dp[index_five]*5;dp[i] = min(min(two, three), five);if (dp[i]==two){index_two++;}if (dp[i]==three){index_three++;}if (dp[i]==five){index_five++;}}return dp[n-1];} };21.不同的二叉搜索樹(shù)
思路:卡塔蘭數(shù)
將 1?(i?1) 序列作為左子樹(shù)，將 (i+1)?n 序列作為右子樹(shù)。接著我們可以按照同樣的方式遞歸構(gòu)建左子樹(shù)和右子樹(shù)。
在上述構(gòu)建的過(guò)程中，由于根的值不同，因此我們能保證每棵二叉搜索樹(shù)是唯一的.也就得到卡塔蘭數(shù)
class Solution(object):def numTrees(self, n):""":type n: int:rtype: int"""#狀態(tài)方程 和G(j-1) * G(n-j)dp = [0]*(n+1)#0 1樹(shù)都為1dp[0], dp[1] = 1, 1for i in range(2,n+1):for j in range(1,i+1):dp[i] += dp[j-1]*dp[i-j]# print('==dp:', dp)return dp[-1]c++實(shí)現(xiàn):
class Solution { public:int numTrees(int n) {vector<int> res(n+1,0); res[0] = 1;res[1] = 1;for (int i=2;i<n+1;i++){for (int j=1;j<i+1;j++){res[i] += res[j-1] * res[i-j];}}return res[n];} };22.?楊輝三角
python:
class Solution:def generate(self, numRows: int) -> List[List[int]]:res = []for i in range(numRows):temp = [0]*(i+1)for j in range(i + 1):if j == 0 or i == j:temp[j] = 1else:temp[j] = res[i-1][j] + res[i-1][j-1]res.append(temp)return resc++:
class Solution { public:vector<vector<int>> generate(int numRows) {vector<vector<int>> res;for(int i = 0; i < numRows; i++){vector<int> temp(i+1, 0);for(int j = 0; j < i+1; j++){if(j == 0 || i == j){temp[j] = 1;}else{temp[j] = res[i-1][j-1] + res[i-1][j];}}res.push_back(temp);}return res;} };?23.?楊輝三角 II
python:?
class Solution:def getRow(self, rowIndex: int) -> List[int]:res = [0]*(rowIndex+1)res[0] = 1for i in range(rowIndex + 1):for j in range(i, 0, -1):res[j] += res[j-1]return res?c++:
class Solution { public:vector<int> getRow(int rowIndex) {vector<int> res(rowIndex+1, 0);res[0] = 1;for(int i = 1; i < rowIndex + 1; i++){for(int j = i; j > 0; j--){res[j] += res[j-1];}}return res;} };24.最小路徑和
思路:動(dòng)態(tài)規(guī)劃?dp[i][j] = min(dp[i-1][j],dp[i][j-1])+v[i][j]
import numpy as np #dp[i][j] = min(dp[i-1][j],dp[i][j-1])+v[i][j] class Solution:def minPathSum(self, grid):h = len(grid)w = len(grid[0])dp = [[0 for i in range(w)] for j in range(h)]dp[0][0] = grid[0][0]for i in range(1, h):dp[i][0] = dp[i-1][0]+grid[i][0]for i in range(1, w):dp[0][i] = dp[0][i-1]+grid[0][i]print('==np.array(dp):\n', np.array(dp))for i in range(1, h):for j in range(1, w):dp[i][j] = min(dp[i-1][j],dp[i][j-1])+grid[i][j]print('==np.array(dp):\n', np.array(dp))return dp[-1][-1] grid = [[1,3,1],[1,5,1],[4,2,1]] sol = Solution() sol.minPathSum(grid)c++:
class Solution { public:int minPathSum(vector<vector<int>>& grid) {//dp[i][j] = min(dp[i-1][j], dp[i][j-1])+matrix[i][j]int m = grid.size();int n = grid[0].size();vector<vector<int>> dp(m, vector<int>(n, 0));dp[0][0] = grid[0][0];for(int i = 1; i < m; i++){dp[i][0] = dp[i-1][0] + grid[i][0];}for(int j = 1; j < n; j++){dp[0][j] = dp[0][j-1] + grid[0][j];}for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];}}return dp[m-1][n-1];} };25.?下降路徑最小和
狀態(tài)轉(zhuǎn)移方程:（注意邊界）?
dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i-1][j+1])+matrix[i][j]
python:
class Solution:def minFallingPathSum(self, matrix: List[List[int]]) -> int:#dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i-1][j+1])+matrix[i][j]n = len(matrix)# dp = [[0 for i in range(n)] for i in range(n)]# for i in range(n):# dp[0][i] = matrix[0][i]for i in range(1, n):for j in range(n):if j == 0:matrix[i][j] = min(matrix[i-1][j], matrix[i-1][j+1])+matrix[i][j] elif j == n-1:matrix[i][j] = min(matrix[i-1][j-1], matrix[i-1][j])+matrix[i][j] else:matrix[i][j] = min(matrix[i-1][j-1], matrix[i-1][j], matrix[i-1][j+1])+matrix[i][j]# print(dp)return min(matrix[-1])c++:
class Solution { public:int minFallingPathSum(vector<vector<int>>& matrix) {//dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i-1][j+1])+matrix[i][j]int n = matrix.size();vector<vector<int>> dp(n, vector<int>(n, 0));for(int i = 0; i < n; i++){dp[0][i] = matrix[0][i];}for(int i = 1; i < n; i++){for(int j = 0; j < n; j++){if(j == 0){dp[i][j] = min(dp[i-1][j], dp[i-1][j+1])+matrix[i][j];}else if(j == n-1){dp[i][j] = min(dp[i-1][j-1], dp[i-1][j])+matrix[i][j];}else{dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i-1][j+1]))+matrix[i][j];}}}int min_value = *min_element(dp[n-1].begin(), dp[n-1].end());return min_value;} };26.三角形最小路徑和
狀態(tài)轉(zhuǎn)移方程，注意邊界?
dp[i][j] = min(dp[i-1][j-1],dp[i-1][j])+ matrix[i][j]
python:
class Solution:def minimumTotal(self, triangle: List[List[int]]) -> int:#dp[i][j] = min(dp[i-1][j-1],dp[i-1][j])+ matrix[i][j]for i in range(1, len(triangle)):for j in range(len(triangle[i])):if j == 0:triangle[i][j] = triangle[i-1][j] + triangle[i][j]elif j == i:triangle[i][j] = triangle[i-1][j-1] + triangle[i][j]else:triangle[i][j] = min(triangle[i-1][j-1], triangle[i-1][j]) + triangle[i][j]return min(triangle[-1])c++:?
class Solution { public:int minimumTotal(vector<vector<int>>& triangle) {int h = triangle.size();for(int i = 1; i < triangle.size(); i++){for(int j = 0; j < triangle[i].size(); j++){if(j == 0){triangle[i][j] = triangle[i-1][j] + triangle[i][j];}else if(j == i){triangle[i][j] = triangle[i-1][j-1] + triangle[i][j];}else{triangle[i][j] = min(triangle[i-1][j-1], triangle[i-1][j]) + triangle[i][j];}}}int min_value = *min_element(triangle[h - 1].begin(), triangle[h - 1].end());return min_value;} };27.?不同路徑
?python:
class Solution:def uniquePaths(self, m: int, n: int) -> int:dp = [[0 for i in range(n)] for j in range(m)]for i in range(m):dp[i][0] = 1for i in range(n):dp[0][i] = 1# print('==np.array(dp):', np.array(dp))for i in range(1,m):for j in range(1,n):dp[i][j] = dp[i-1][j]+dp[i][j-1]# print(np.array(dp))return dp[-1][-1]c++:
class Solution { public:int uniquePaths(int m, int n) {vector<vector<int>> dp(m, vector<int>(n, 0));for(int i = 0; i < m; i++){dp[i][0] = 1;}for(int j = 1; j < n; j++){dp[0][j] = 1;}for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){dp[i][j] = dp[i-1][j] + dp[i][j-1];}}return dp[m-1][n-1];} };28.不同路徑 II
?
?python:
class Solution:def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:if obstacleGrid[0][0] == 1:return 0m, n = len(obstacleGrid), len(obstacleGrid[0])dp = [[0 for _ in range(n)] for _ in range(m)]dp[0][0] = 1for i in range(1, m):if dp[i-1][0] == 1 and obstacleGrid[i][0] != 1:dp[i][0] = 1for j in range(1, n):if dp[0][j-1] ==1 and obstacleGrid[0][j] != 1:dp[0][j] = 1for i in range(1, m):for j in range(1, n):if obstacleGrid[i][j] != 1:dp[i][j] = dp[i-1][j] + dp[i][j-1]return dp[m-1][n-1]c++:
class Solution { public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m = obstacleGrid.size();int n = obstacleGrid[0].size();vector<vector<int>> dp(m, vector<int>(n, 0));if(obstacleGrid[0][0] == 1){return 0;}dp[0][0] = 1;for(int i = 1; i < m; i++){if(dp[i-1][0]==1 && obstacleGrid[i][0] != 1){dp[i][0] = 1;} }for(int j = 1; j < n; j++){if(dp[0][j-1] == 1 && obstacleGrid[0][j] != 1){dp[0][j] = 1;} }for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){if(obstacleGrid[i][j] != 1){dp[i][j] = dp[i-1][j] + dp[i][j-1];}}}return dp[m-1][n-1];} };29.最大正方形
python:
狀態(tài)轉(zhuǎn)移方程
matrix[i][j] == '1'
dp[i][j] =?min(dp[i-1][j], dp[i-1][j-1], dp[i][j-1]) + 1
class Solution:def maximalSquare(self, matrix: List[List[str]]) -> int:h = len(matrix)w = len(matrix[0])dp = [[0 for j in range(w)] for i in range(h)]dp[0][0] = int(matrix[0][0])maxside = int(matrix[0][0])for i in range(1, h):dp[i][0] = int(matrix[i][0])maxside = max(maxside, dp[i][0]) for j in range(1, w):dp[0][j] = int(matrix[0][j])maxside = max(maxside, dp[0][j])for i in range(1, h):for j in range(1, w):if matrix[i][j] == '1':dp[i][j] = min(dp[i-1][j], dp[i-1][j-1], dp[i][j-1]) + 1 maxside = max(maxside, dp[i][j])return maxside**2c++:
class Solution { public:int maximalSquare(vector<vector<char>>& matrix) {int m = matrix.size();int n = matrix[0].size();vector<vector<int>> dp(m, vector<int>(n, 0));int maxside = matrix[0][0] - '0';dp[0][0] = matrix[0][0] - '0';for(int i = 1; i < m; i++){dp[i][0] = matrix[i][0] - '0';maxside = max(maxside, dp[i][0]);}for(int j = 1; j < n; j++){dp[0][j] = matrix[0][j] - '0';maxside = max(maxside, dp[0][j]);}// cout<<maxside<<endl;for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){if(matrix[i][j] == '1'){dp[i][j] = min(dp[i][j-1], min(dp[i-1][j-1], dp[i-1][j])) + 1;maxside = max(maxside, dp[i][j]);}}}return maxside*maxside;}};30.最長(zhǎng)回文子串
思路:中心擴(kuò)散或者動(dòng)態(tài)規(guī)劃
class Solution:def helper(self,left,right,s):while left>=0 and right<len(s) and s[left]==s[right]:left-=1right+=1if len(s[left+1:right])>len(self.res):self.res = s[left+1:right]def longestPalindrome(self, s: str) -> str:self.res = ''for i in range(len(s)):self.helper(i,i,s)self.helper(i,i+1,s)return self.res?c++:
class Solution { public:string res;void help(string s, int left, int right){while(left>=0 && right<s.size() && s[left]==s[right]){left--;right++;}left++;right--;if((right - left + 1) > res.size()){res = s.substr(left, right - left + 1); }}string longestPalindrome(string s) {if(s.size()<=1){return s;}for(int i=0; i<s.size(); i++){help(s, i, i);help(s, i, i+1);}return res;} };動(dòng)態(tài)規(guī)劃：
class Solution:def longestPalindrome(self, s: str) -> str:size = len(s)# 特殊處理if size == 1:return s# 創(chuàng)建動(dòng)態(tài)規(guī)劃dynamic programing表dp = [[False for _ in range(size)] for _ in range(size)]# 初始長(zhǎng)度為1，這樣萬(wàn)一不存在回文，就返回第一個(gè)值（初始條件設(shè)置的時(shí)候一定要考慮輸出）max_len = 1start = 0for j in range(1,size):for i in range(j):# 邊界條件：# 只要頭尾相等（s[i]==s[j]）就能返回Trueif j-i<=2:if s[i]==s[j]:dp[i][j] = Truecur_len = j-i+1# 狀態(tài)轉(zhuǎn)移方程 # 當(dāng)前dp[i][j]狀態(tài)：頭尾相等（s[i]==s[j]）# 過(guò)去dp[i][j]狀態(tài)：去掉頭尾之后還是一個(gè)回文（dp[i+1][j-1] is True）else:if s[i]==s[j] and dp[i+1][j-1]:dp[i][j] = Truecur_len = j-i+1# 出現(xiàn)回文更新輸出if dp[i][j]:if cur_len > max_len:max_len = cur_lenstart = ireturn s[start:start+max_len]31.最長(zhǎng)回文子序列
思路:動(dòng)態(tài)規(guī)劃
class Solution:def longestPalindromeSubseq(self, s: str) -> int:n = len(s)dp = [[0] * n for _ in range(n)]for i in range(n - 1, -1, -1):dp[i][i] = 1for j in range(i + 1, n):if s[i] == s[j]:dp[i][j] = dp[i + 1][j - 1] + 2else:dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])return dp[0][n - 1]32.最長(zhǎng)遞增子序列
思路:動(dòng)態(tài)規(guī)劃
python:
class Solution:def lengthOfLIS(self, nums: List[int]) -> int:if len(nums)==0:return 0opt = [1]*len(nums)for i in range(1, len(nums)):for j in range(i):if nums[i]>nums[j]:opt[i] = max(opt[i], opt[j]+1)# print('==value:', value)# opt[i] = value+1# print('==dp:', opt)return max(opt)c++:
class Solution { public:int lengthOfLIS(vector<int>& nums) {int n = nums.size();vector<int> dp(n, 1);int max_value = 1;for(int i = 1; i < n; i++){for(int j = 0; j < i; j++){if(nums[j] < nums[i]){dp[i] = max(dp[i], dp[j] + 1);max_value = max(dp[i], max_value);}}}return max_value;} };33.判斷子序列
思路：雙指針
class Solution:def isSubsequence(self, s: str, t: str) -> bool:#雙指針i, j = 0, 0while i < len(s) and j < len(t):if s[i] == t[j]:i += 1j += 1else:j += 1if i == len(s):return Trueelse:return False class Solution { public:bool isSubsequence(string s, string t) {int i=0,j=0;while(i < s.size() && j < t.size()){if(s[i] == t[j]){i++;j++;}else{j++;}}if(i == s.size()){return true;}else{return false;}} };34-1.最長(zhǎng)公共子序列
python:?
class Solution:def minDistance(self, word1: str, word2: str) -> int:m = len(word1)n = len(word2)dp = [[0 for i in range(n+1)] for i in range(m+1)]for i in range(1, m+1):dp[i][0] = ifor j in range(1, n+1):dp[0][j] = jfor i in range(m):for j in range(n):if word1[i] == word2[j]:dp[i+1][j+1] = dp[i][j]else:dp[i+1][j+1] = min(dp[i][j], dp[i+1][j], dp[i][j+1]) + 1return dp[m][n]c++:?
class Solution { public:int longestCommonSubsequence(string text1, string text2) {int m = text1.size();int n = text2.size();vector<vector<int>> dp(m+1, vector<int>(n+1, 0));for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){if(text1[i] == text2[j]){dp[i+1][j+1] = dp[i][j] + 1;}else{dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);}}}return dp[m][n];} };34-2.最長(zhǎng)公共字串
# # 動(dòng)態(tài)規(guī)劃解決最大公共子串問(wèn)題 def find_lcsubstr(s1, s2):m = [[0 for i in range(len(s2) + 1)] for j in range(len(s1) + 1)] # 生成0矩陣，為方便后續(xù)計(jì)算，比字符串長(zhǎng)度多了一列print(m)mmax = 0 # 最長(zhǎng)匹配的長(zhǎng)度p = 0 # 最長(zhǎng)匹配對(duì)應(yīng)在si中的最后一位for i in range(len(s1)):for j in range(len(s2)):if s1[i] == s2[j]:m[i + 1][j + 1] = m[i][j] + 1if m[i + 1][j + 1] > mmax:mmax = m[i + 1][j + 1]p = i + 1print(p)return s1[(p - mmax):p], mmax # 返回最長(zhǎng)子串及其長(zhǎng)度?35.編輯距離
?編輯距離，又稱Levenshtein距離（萊文斯坦距離也叫做Edit Distance），是指兩個(gè)字串之間，由一個(gè)轉(zhuǎn)成另一個(gè)所需的最少編輯操作次數(shù)，如果它們的距離越大，說(shuō)明它們?cè)绞遣煌ＴS可的編輯操作包括將一個(gè)字符替換成另一個(gè)字符，插入一個(gè)字符，刪除一個(gè)字符。
mat[i+1,j]+1表示增加操作
d[i,j+1]+1 表示刪除操作
d[i,j]+temp表示替換操作,其中temp取0或1
c++實(shí)現(xiàn):
class Solution { public:int minDistance(string word1, string word2) {int h = word1.size();int w = word2.size();vector<vector<int>> opt(h + 1, vector<int>(w + 1, 0));for(int i = 0; i < h; i++){opt[i + 1][0] = i + 1;}for(int j = 0; j < w; j++){opt[0][j + 1] = j + 1;}for(int i = 0; i < h; i++){for (int j = 0; j < w; j++){if(word1[i] == word2[j]){opt[i + 1][j + 1] = opt[i][j];}else{opt[i + 1][j + 1] = min(opt[i][j] + 1, min(opt[i + 1][j] + 1, opt[i][j + 1] + 1));}}}return opt[h][w];} };36.零錢兌換
?思路：找準(zhǔn)狀態(tài)狀轉(zhuǎn)移方程,f代表選擇銀幣的函數(shù)，則f(11)=f(11-1)+1或f(11)=f(11-2)+1或f(11)=f(11-5)+1,則一般方程為:
f(money) = min(f(money), f(money-coin)+1)
class Solution:def coinChange(self, coins: List[int], amount: int) -> int:#狀態(tài)轉(zhuǎn)移方程f(money) = min(f(money),f(money-coin)+1)f = [float('inf')] * (amount + 1)f[0] = 0# print('==f:', f)for i in range(1, amount + 1):for coin in coins:if i - coin >= 0:f[i] = min(f[i], f[i - coin] + 1)# print('==f:', f)return f[-1] if f[-1]!=float('inf') else -1c++實(shí)現(xiàn)：
class Solution { public:int coinChange(vector<int>& coins, int amount) {vector<int> dp(amount+1, INT_MAX-1);dp[0] = 0;for(int i = 1; i < amount + 1; i++){for(int j = 0; j < coins.size(); j++){if(i - coins[j] >= 0){dp[i] = min(dp[i - coins[j]] + 1, dp[i]);}}}if(dp[amount] == INT_MAX-1){return -1;}else{return dp[amount];}} };3?????????????7:零錢兌換 II
思路1:回溯 會(huì)超時(shí)
# 組合問(wèn)題 回溯 超時(shí) class Solution:def backtrace(self, amount, start, coins, track):if amount == 0: # 終止條件# self.res.append(track)self.res+=1returnfor i in range(start, len(coins)): # 選擇條件if coins[i] > amount:continue# store = track.copy()# track.append(coins[i])self.backtrace(amount - coins[i], i, coins, track)# track = storedef change(self, amount, coins):self.res = 0#[]coins = sorted(coins)self.backtrace(amount, 0, coins, [])return self.res# amount = 5 # coins = [2] amount = 5 coins = [1, 2, 5] # amount = 500 # coins = [3,5,7,8,9,10,11] sol = Solution() res = sol.change(amount, coins) print('==res:', res)思路２:當(dāng)成完全背包問(wèn)題，用dp
#dp[i][j] 硬幣為i 金額為j的組合數(shù) import numpy as np class Solution:def change(self, amount, coins):if len(coins) == 0:if amount == 0:return 1else:return 0dp = [[0 for i in range(amount+1)] for j in range(len(coins))]print('==np.array(dp):', np.array(dp))dp[0][0] = 1for j in range(coins[0], amount+1, coins[0]):dp[0][j] = 1print('==np.array(dp):', np.array(dp))for i in range(1, len(coins)):print('==coins[i]:', coins[i])for j in range(amount+1):dp[i][j] = dp[i - 1][j]#不選if j >= coins[i]:#選 注意與0 1背包有一點(diǎn)不同dp[i][j] += dp[i][j - coins[i]]print('==np.array(dp):', np.array(dp))return dp[-1][-1]amount = 5 coins = [1, 2, 5] sol = Solution() sol.change(amount, coins)c++實(shí)現(xiàn)：
class Solution { public:int change(int amount, vector<int>& coins) {int m = coins.size();vector<vector<int>> dp(m, vector<int>(amount + 1, 0));dp[0][0] = 1;for(int j = coins[0]; j < amount + 1; j += coins[0]){dp[0][j] = 1;}for(int i = 1; i < m; i++){for(int j = 0; j < amount + 1; j++){dp[i][j] = dp[i-1][j];if(j >= coins[i]){dp[i][j] += dp[i][j - coins[i]];}}}return dp[m-1][amount];} };總結(jié)
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