昨天第一次參加LeetCode Weekly Contest， 一道題沒(méi)有做出來(lái)。所有時(shí)間都花在第一道題上了，被虐得很慘。 看了一下別人的參考代碼，理解之后發(fā)現(xiàn)真的很簡(jiǎn)單。

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 to1 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].

首先在Array里面找到逆序的元素，也就是nums[i] > nums[i + 1], 用reversOrder來(lái)記錄逆序的個(gè)數(shù)。如果個(gè)數(shù)超過(guò)1，則不可能通過(guò)改一個(gè)元素就變成正序，所以返回false; 如果目前是第一個(gè)逆序，則討論兩種情況：

• nums[i - 1] <= nums[i + 1],比如[1,6,2,3,4]里6 > 2 并且5 > 2這種情況，要改正的話將6改成1或者2能滿足題意；但如果寫(xiě)6改成2就可以同時(shí)包括i == 0的情況；
• nums[i - 1] > nums[i + 1], 比如[5,6,2,3,4]里6 > 2并且 5 > 2這種情況，肯定是改2為6才能將此處的逆序改為正序。但這個(gè)例子的情況是改為[5,6,6,3,4]之后仍然有逆序，下次循環(huán)reverseOrder != 0也會(huì)返回false.

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