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
类比LCS问题,但只需要检查前两个点,因此是O(2*n)的复杂度。
特殊样例
{2, 3, 3, 1, 4 }
一开始没注意,这个样例没过去。。。
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;
}
};