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: TrueExplanation: You could modify the first 4 to 1 to get a non-decreasing array.

 

Example 2:

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

 

Note: The n belongs to [1, 10,000].

/*一开始想的有点简单，直接只是判断有多少个不符合的位置，并没有修复并检查修复后是否满足条件  所以 出现  3, 4, 2, 3的case就不行了其实主要分两种情况：2， 4， 2， 33， 4， 2， 3判断当前的值是否比前一个值小，如果小的话， 再判断是否比 前前一个值 要小， 如果小，改变当前的值 使得其跟前一个值一样大，如果大，改变前一个值，使得它跟当前的值一样大。*/class Solution {public:    bool checkPossibility(vector
     
      & nums) {        int len = nums.size();        int count = 0;       // if (len == 0)   return false;        for (int i = 1; i < len; i++){            if (nums[i] < nums[i - 1]){                count ++;                if (count > 1){                    return false;                }                if (nums[i] < nums[i - 2] && (i - 2) >= 0){                    nums[i] = nums[i - 1];                }                else{                    nums[i - 1] = nums[i];                }            }                    }        return true;    }};