leetcode 1186

问题

Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.

Note that the subarray needs to be non-empty after deleting one element.

Example 1:

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Input: arr = [1,-2,0,3]
Output: 4
Explanation: Because we can choose [1, -2, 0, 3] and drop -2, thus the subarray [1, 0, 3] becomes the maximum value.

Example 2:

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Input: arr = [1,-2,-2,3]
Output: 3
Explanation: We just choose [3] and it's the maximum sum.

Example 3:

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Input: arr = [-1,-1,-1,-1]
Output: -1
Explanation: The final subarray needs to be non-empty. You can't choose [-1] and delete -1 from it, then get an empty subarray to make the sum equals to 0.

Constraints:

  • 1 <= arr.length <= 10^5
  • -10^4 <= arr[i] <= 10^4

分析

$dp[i][0]$ 表示没有删除元素且以 $a[i]$ 结尾的最大和的子数组的和。$dp[i][0]$ 表示删除了一个元素且以 $a[i]$（$a[i]$ 也有可能被删除）结尾的最大和的子数组的和。

代码

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class Solution {
public:
    int maximumSum(vector<int>& arr) {
        int n = arr.size();
        vector<int> dp(2);
        dp[0] = arr[0];
        dp[1] = 0;
        int ans = arr[0];  // 注意最后要求的返回数组不能是空
        for (int i = 1; i < n; ++i) {
            dp[1] = max(dp[0], dp[1]+arr[i]);
            dp[0] = max(arr[i], dp[0]+arr[i]);
            ans = max(ans, max(dp[0], dp[1]));
        }
        return ans;
    }
};