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
類比LCS問題，但只需要檢查前兩個點，因此是O（2*n）的複雜度。
特殊樣例
{2, 3, 3, 1, 4 }
一開始沒注意，這個樣例沒過去。。。

class Solution {
public:
    bool checkPossibility(vector<
 
int>& arr) {
        int n=arr.size();
        int dp[100000];
        fill(dp, dp+n, 1);
        for(int i=1;i<n;i++) {
            if(i-1>=0&&arr[i-1]<=arr[i]) 
                dp[i]=dp[i-1]+1;
            if(i-2>=0&&arr[i-2]<=arr[i])
                dp[i]=max(dp[i],
 
 dp[i-2]+1);
        }
        return dp[n-1]>=n-1||dp[n-2]>=n-1;
    } 
};