leetcode先刷_Maximum Subarray

dp创始人级精英赛的冠军。最大的部分和。

扫从左至右，保持一个最佳值而当前部分和，在这一部分，并成为负值什么时候。再往下的积累后，也起到了负面作用，所以，放弃直销，然后部分和初始化为阅读的当前位置。

class Solution {
public:int maxSubArray(int A[], int n) {int mmax = A[0], tpsum = A[0];for(int i=1;i<n;i++){if(tpsum<0)     tpsum = A[i];else    tpsum += A[i];if(tpsum > mmax)mmax = tpsum;}return mmax;}
};

转载于:https://www.cnblogs.com/yxwkf/p/4754488.html

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