算法第16周Maximum Subarray[easy]

Description

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

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Solution

这道题是求最大连续子串的和。我们可以采用动态规划的方法来解决。
dp[i] 为以A[i]结尾的最大连续子串的和。
如果dp[i-1] < 0, 那么以A[i]结尾的话，无论A[i]为正还是负，加上一个负数，总会小于它本身，所以dp[i] = A[i];
如果dp[i-1] > 0, 那么dp[i] = A[i]+dp[i-1];
在这个过程中需要记录dp数组的最大值。

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