[LeetCode] 665. Non-decreasing Array

标签：判断   des   efi   put   用例   ast   integer   pre   遇到   

Description

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].

Analyse

给定一个大小为n的数组A，允许最多修改一个元素的值，判断能否变成一个非递减序列

从左到右遍历数组，当遇到比前一个数A[i-1]小的数A[i]的时候，由于可以修改一个元素的值，可以将前一个数A[i-1]变小，A[i-1] = A[i-2]，

反映到代码里，直接判断当前数A[i]是否大于A[i-2]即可

有一种不满足以上关系的情况(1-2<0)

A[1] < A[0]

比如{4, 2, 3}，这种情况就直接忽略4，判断剩下的元素是否非递减的

最终代码如下，这道题我提交了快10次，拿到了8个LeetCode的测试用例，最后看着Solution改自己的版本才AC了

Think more, write less

bool checkPossibility(vector<int>& nums)
{
    int last = -100000;
    bool state = false;
    for (int i = 0; i < nums.size(); i++)
    {
        if (nums[i] >= last)
        {
            last = nums[i];
        }
        else
        {
            if (state) return false;

            if (i==1 || nums[i] >= nums[i-2])
            {
                last = nums[i];
            }

            state = true;
        }
    }

    return true;
}

[LeetCode] 665. Non-decreasing Array

标签：判断   des   efi   put   用例   ast   integer   pre   遇到   

原文地址：https://www.cnblogs.com/arcsinw/p/10661968.html