剑指 Offer 42. 连续子数组的最大和

public int maxSubArray(int[] nums) {
        int n = nums.length;
        int[] dp = new int[n+1];
        int ans = Integer.MIN_VALUE;
        for(int i = 1 ; i<= n ; i++)
        {
            dp[i] = Math.max(dp[i-1] + nums[i-1], nums[i-1]);
            ans = Math.max(ans,dp[i]);

        }
        return ans;
    }

 

posted @ 2020-07-13 09:43  贼心~不死  阅读(92)  评论(0编辑  收藏  举报