# 665.Non-Decreasing Array

**665.Non-Decreasing Array**

难度:Easy

> 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].
```

找出第一个非递减数列的位置，将数组分为两个子数组，如果两个子数组都是有序的，则有可能通过修改一个元素来满足条件。如果子数组中并不是非递减的，则不可能通过只修改一个元素来满足非递减数组。 对于不满足条件的两个元素，分别尝试通过修改这两个元素是否能满足条件。

```
class Solution {
public:
    bool checkPossibility(vector<int>& nums) {
        if(nums.size()<3) return true;
        int index=nums.size();
        for(int i=1;i<nums.size();i++)
        {
            if(nums[i-1] >nums[i] )
            {
                index=i;
                break;
            }
        }
        if( index>=nums.size()-1) return true;
        for(int i=index+1;i<nums.size();i++)
        {
            if(nums[i]<nums[i-1]) return false;
        }
        if(index ==1 ) return true;
            //cout<<index<<endl;
        if( nums[index-2]<=nums[index]) return true;
        if( nums[index-1]<=nums[index+1]) return true;
    
        return false;
    }
};
```

> 执行用时 :32 ms, 在所有 C++ 提交中击败了89.45%的用户\
> 内存消耗 :10.3 MB, 在所有 C++ 提交中击败了84.21%的用户


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