[Leetcode] 665. Non-decreasing Array 解题报告_delete l到r decreasing csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79189354

本文介绍一种算法，用于判断一个整数数组通过修改最多一个元素能否变为非递减序列。文中提供了具体的实现思路及C++代码示例。

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题目：

Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.

We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n).

Example 1:

Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.

Example 2:

Input: [4,2,1]
Output: False
Explanation: You can't get a non-decreasing array by modify at most one element.

Note: The n belongs to [1, 10,000].

思路：

我们nums[i]和其之前的数字nums[i - 1]，如果nums[i] < nums[i - 1]，那么有两种情况：1）如果nums[i - 2] <= nums[i]，那么说明nums[i - 1]是最大的，我们只需要将其删除就可以消除当前的递减情况（在实现中，我们通过修改nums[i - 1]的值来实现）；2）如果nums[i - 2] > nums[i]，那么nums[i]是最小的，我们需要删除掉nums[i]才有可能消除递减情况（在实现中我们通过修改nums[i]的值来实现）。

代码：

class Solution {
public:
    bool checkPossibility(vector<int>& nums) {
        int count = 0;
        for (int i = 1; i < nums.size() && count <= 1; ++i) {
            if (nums[i] < nums[i - 1]) {                // nums[i - 1] is larger
                ++count;
                if (i < 2 || nums[i - 2] <= nums[i]) {
                    nums[i - 1] = nums[i];              // we want to delete nums[i - 1]
                }
                else {                                  // i >= 2 && nums[i - 2] > nums[i]
                    nums[i] = nums[i - 1];              // we want to delete nums[i]
                }
            }
        }
        return count <= 1;
    }
};