665. Non-decreasing Array

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].

 

class Solution {
    public boolean checkPossibility(int[] nums) {
        if (nums == null || nums.length < 3) {
            return true;
        }
        int cnt = 0;
        for (int i = 1; i< nums.length; i++) {
            if (nums[i] < nums[i - 1]) {
                if (i < nums.length - 1 && nums[i + 1] < nums[i - 1]) {
                    nums[i - 1] = nums[i];
                    cnt++;
                    if (i - 2 >= 0 && nums[i - 2] > nums[i - 1]) {
                        return false;
                    }
                } else {
                    nums[i] = nums[i - 1];
                    cnt++;
                }
            }
            if (cnt > 1) {
                return false;
            }
        }
        return true;
    }
}

　　

　　

We can also do it without modifying the input by using a variable prev to hold the a[i-1]; if we have to lower a[i] to match a[i-1] instead of raising a[i-1], simply skip updating prev;

without modified

class Solution {
    public boolean checkPossibility(int[] a) {
        int modified = 0;
        for (int i = 1, prev = a[0]; i < a.length; i++) {
            if (a[i] < prev) {
                if (modified++ > 0) return false;
                if (i - 2 >= 0 && a[i - 2] > a[i]) continue;
            }
            prev = a[i];
        }
        return true;
    }
}