[leetcode-665-Non-decreasing Array]-优快云博客

本文介绍了一个算法问题的解决方案，即判断一个整数数组通过修改最多一个元素是否可以变成非递减序列。文章提供了详细的逻辑分析及C++实现代码。

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：

考虑数组中逆序发生的次数，如果cnt>=2，返回false。如果cnt==1的话需要判断一下。

比如 xxx 5 3 7 xx 发生了一次逆序 5比7小只需将3调整为7就可以了
还有就是比如 xxxxx2725xxx 从7那下降了然后从2后上升了但是此时把7改成2就可以啦
这个就是属于if (nums[index - 1] <= nums[index + 1])return true;

就是先下降后上升如果上升的更高那就改一次就行了

但是如果没更高那就试试把左边的也下降一下看满足条件不如果满足就true 否则false

bool checkPossibility(vector<int>& nums)
{
    if (nums.size() <=2)return true;
    int  cnt = 0,index =0;
    for (int i = 0; i < nums.size()-1;i++)
    {
        if (nums[i]>nums[i + 1])
        {
            cnt++;
            index = i;//逆序发生位置
        }
    }    
    if (cnt == 0)return true;
    if (cnt == 1)
    {
        if (index == 0 || index == nums.size() - 1)return true;
        if (nums[index] <= nums[index + 2])return true;
        if (nums[index - 1] <= nums[index + 1])return true;
        return false;
    }    
    return false;
}