Description Link to heading

53.maximum-subarray

Solution Link to heading

dp[i] denotes maximum sum of subarray ending with nums[i - 1], then: dp[i] = max(dp[i - 1] + nums[i - 1], nums[i - 1]);

Code Link to heading 

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        vector<int> dp(nums.size() + 1, 0);
        dp[1] = nums[0];
        int m = nums[0];
        for (int i = 1; i <= nums.size(); i++) {
            dp[i] = max(dp[i - 1] + nums[i - 1], nums[i - 1]);
            if (dp[i] > m)
                m = dp[i];
        }
        return m;
    }
};