Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],Output: 6Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

public class MaximumSubarray{    public static int maximuntmSubarray(int[] a){        int n = n.length;        int max = a[0];        int[] dp = new int[n];        dp[0] = a[0];        for(int i = 1; i < n; i++){            dp[i] = (dp[i-1]>0 ? dp[i] : 0)+a[i];            max = Math.max(max,dp[i]);        }        return max;    }}

 

 

经典的动态规划问题。令数组dp[i]为前i个数的最大连续数之和，max初始值为a[0]。分解问题：

当前i-1个数的最大连续和为负时，那么另dp[i]=a[i];

当前i-1个数的最大连续和不为负时，另dp[i] = a[i] + dp[i-1];

最后与保存的max相比较，取最大值。