Leetcode brush first_ Maximum Subarray

Hello everyone , I meet you again , I'm the king of the whole stack .

dp The champion of Founder level elite . The largest part and .

Sweep from left to right , Maintain an optimal value while the current part and , In this part , And become negative when . After further accumulation , Also played a negative role , therefore , Abandon direct selling , Then partial sum is initialized to the current position of reading .

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;
    }
};

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