題目：

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].

解釋：
判斷陣列是否能在改變最多一個數字的情況下變成非遞減序列。
當遇到nums[i]>nums[i+1]的情況，我們有兩種選擇使得陣列非遞減（區域性）：
1.把nums[i]降低為nums[i+1]
2.把nums[i+1]升高為nums[i]
如果可0.行的話，當然是選擇優先把 nums[i]

class Solution(object):
    def checkPossibility
 
(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        modified=False
        for i in range(len(nums)-1):
            if nums[i]>nums[i+1]:
                if modified:
                    return False
                if i<1 or nums[i+1]>nums[i-1]:
                    nums[i]=nums[i+1]
                else:
                    nums[i+1]=nums[i]
                modified=True
        return True

c++程式碼：

class Solution {
public:
    bool checkPossibility(vector<int>& nums) {
        bool modified=false;
        for (int i=0;i<nums.size()-1;i++)
        {
            if (nums[i]>nums[i+1])
            {
                if(modified)
                    return false;
                if(i<1 || nums[i-1]<nums[i+1])
                    nums[i]=nums[i+1];
                else
                    nums[i+1]=nums[i];
                modified=true;
            }
        }
        return true;
    }
};

總結：