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].

自己琢磨的 :(简直无法忍受)

class Solution {
     public boolean checkPossibility(int[] nums) {
       if(nums.length<=1){
			return true;
		}
		
		List<Integer> list = new ArrayList<>();
        for(int i = 0;i<nums.length-1;i++){
           if(nums[i]>nums[i+1]){
        	   list.add(i+1);
           }
        }
        if(list.size()>1){
        	return false;
        }
        if(list.size()== 0 || list.get(0)==1 || list.get(0)==nums.length-1){
        	return true;
        }
        int left =0;
        for(int i = 0;i<list.get(0);i++){
        	if(nums[i]>nums[list.get(0)]){
        		left++;
        	}
        }
        int right = 0;
        for (int i = list.get(0); i < nums.length; i++) {
        	if(nums[i]<nums[list.get(0)-1]){
        		right++;
        	}
		}
        
        if(left >1 && right>1){
    		return false;
        }
        return true;
    }
}

大佬的答案:

class Solution {
    public boolean checkPossibility(int[] nums) {
       int count = 0;
        for(int i = 0;i<nums.length -1 ;i++){
            if(nums[i]>nums[i+1]){
                count++;
                if(i>0 && nums[i+1]<nums[i-1]){
                    nums[i+1] = nums[i];
                }
            }
        }
        return count<=1;
    }
}