[LeetCode] Non-decreasing Array

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].

给定一个包含n个数的数组，判断如果最多修改数组中的1个数字，使其变成一个非减序列。

首先找出不符合条件的数组。这样的数组有2种情况：

一，形如[4, 2, 1]的情况，连续相邻的2个数，前一个数都比后一个数大。返回false

二，形如[3, 4, 2, 3]的情况，第一个数比第三个数大，第二个数比第四个数大。返回false （需要注意i的取值情况）

遍历数组时对这两种情况进行判断即可。

class Solution {
public:
    bool checkPossibility(vector<int>& nums) {
        int n = 0;
        for (int i = 1; i != nums.size(); i++) {
            if (nums[i - 1] > nums[i]) {
                n++;
                if (n == 2)
                    return false;
            }
            if (i != 1 && i != nums.size() - 1)
                if (nums[i - 2] > nums[i] && nums[i - 1] > nums[i + 1])
                    return false;
        }
        return true;
    }
};
// 39 ms