LeetCode45 Jump Game II
題目:
Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
For example:
Given array A =?[2,3,1,1,4]
The minimum number of jumps to reach the last index is?2. (Jump?1?step from index 0 to 1, then?3?steps to the last index.)
Note:
You can assume that you can always reach the last index. (Hard)
分析:
第一眼看完題目感覺用動(dòng)態(tài)規(guī)劃肯定能解,但是感覺就是有很多重復(fù)計(jì)算在里面...因?yàn)轭}目只問了個(gè)最少步數(shù)。
果然寫出來(lái)之后華麗超時(shí),但是代碼還是記錄一下吧
1 class Solution { 2 public: 3 int jump(vector<int>& nums) { 4 int dp[nums.size()]; 5 for (int i = 0; i < nums.size(); ++i) { 6 dp[i] = 0x7FFFFFFF; 7 } 8 dp[0] = 0; 9 for (int i = 1; i < nums.size(); ++i) { 10 for (int j = 0; j < i; ++j) { 11 if (j + nums[j] >= i) { 12 dp[i] = min(dp[i], dp[j] + 1); 13 } 14 } 15 } 16 return dp[nums.size() - 1]; 17 } 18 };優(yōu)化考慮類似BFS的想法,維護(hù)一個(gè)步數(shù)的范圍和一個(gè)能走到的最遠(yuǎn)距離。
算法就是遍歷一遍數(shù)組,到每個(gè)位置的時(shí)候,更新他能到達(dá)的最遠(yuǎn)距離end,如果一旦超過(guò)nums.size() - 1,就返回step + 1;
curEnd維護(hù)以當(dāng)前步數(shù)能到達(dá)的最遠(yuǎn)范圍,所以當(dāng)i > curEnd時(shí),step++,并且將curEnd更新為end。
代碼:
1 class Solution { 2 public: 3 int jump(vector<int>& nums) { 4 if (nums.size() == 1) { 5 return 0; 6 } 7 int step = 0, end = 0, curEnd = 0; 8 for (int i = 0; i < nums.size(); ++i) { 9 if (i > curEnd) { 10 step++; 11 curEnd = end; 12 } 13 end = max(end, nums[i] + i); 14 if (end >= nums.size() - 1) { 15 return step + 1; 16 } 17 } 18 return -1; 19 } 20 };?
轉(zhuǎn)載于:https://www.cnblogs.com/wangxiaobao/p/5835856.html
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