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題意：尋找子數(shù)組的和最大。

思路：設(shè)置dp數(shù)組來(lái)保存到第i位的最大和。

判斷第i-1位的正負(fù)，若dp[i-1]<0 則?dp[i] = nums[i]; 若 dp[i-1] > 0 則?dp[i] = dp[i-1] +nums[i];

最后用?max_num = max(max_num, dp[i]); 來(lái)存儲(chǔ)最大的和。

class Solution { public:int maxSubArray(vector<int>& nums) {int len = nums.size();if(len == 0) return 0;if(len == 1) return nums[0];vector<int> dp(len, 0); //dp[i]: 以第i個(gè)元素為結(jié)尾的最大子序列和dp[0] = nums[0];int max_num = dp[0];for(int i=1; i<len; i++){if(dp[i-1] > 0)dp[i] = dp[i-1] +nums[i];elsedp[i] = nums[i];}return max_num;} };

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題意：求子數(shù)組（數(shù)組里的數(shù)是連續(xù)）的最大乘積。

思路：因?yàn)閿?shù)組里有負(fù)號(hào)，所以用兩個(gè)數(shù)組dp_max 和 dp_min 分別保存到當(dāng)前位置i的最大值和最小值。

狀態(tài)轉(zhuǎn)移方程：???????????

dp_max[i%2] = max( max(dp_max[(i-1)%2]*nums[i], nums[i]), dp_min[(i-1)%2]*nums[i]);
dp_min[i%2] = min(min(dp_min[(i-1)%2]*nums[i], nums[i]), dp_max[(i-1)%2]*nums[i]);

class Solution { public:int maxProduct(vector<int>& nums) {int n = nums.size();if(n==0) return 0;int dp_max[2]={nums[0], 0}, dp_min[2]={nums[0], 0} ;int ans = nums[0];for(int i=1; i<n; i++){dp_max[i%2] = max( max(dp_max[(i-1)%2]*nums[i], nums[i]), dp_min[(i-1)%2]*nums[i]);dp_min[i%2] = min(min(dp_min[(i-1)%2]*nums[i], nums[i]), dp_max[(i-1)%2]*nums[i]);ans = max(ans, dp_max[i%2]);}return ans;} };

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266. Palindrome Permutation 回文全排列

Given a string, determine if a permutation of the string could form a palindrome.

Example 1:

Input: "code" Output: false

Example 2:

Input: "aab" Output: true

Example 3:

Input: "carerac" Output: true

Hint:

Consider the palindromes of odd vs even length. What difference do you notice?

Count the frequency of each character.

If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times?

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轉(zhuǎn)載于:https://www.cnblogs.com/Bella2017/p/10957868.html

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