Array-Non-decreasing Array

最新推荐文章于 2025-08-14 11:49:44 发布

kevin聪

最新推荐文章于 2025-08-14 11:49:44 发布

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分类专栏：算法与数据结构 leetcode 文章标签：算法数据结构 leetcode

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本文介绍了一个算法问题：通过修改数组中的最多一个元素来判断数组是否能变为非递减序列。提供了详细的解决方案及代码实现。

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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].

Solution:

//可以这样理解,在a[i]之前为一条递增的直线,a[i]这里突然下降了,那么为了消除这个下降就有两个选择,一个是忽略掉a[i],还是以原来的直线递增,一个是忽略掉a[i-1]，相当于直线朝a[i]折了一下,然后继续判断，如果下次有点下降，那么返回false。那么如何做出选择呢?如果a[i] < a[i - 1]&&a[i] < a[i - 2],那么忽略a[i](因为若忽略a[i-1],直线扔有下降部分),如果a[i]<a[i - 1]&&a[i]>=a[i - 2],那么忽略a[i - 1](因为若忽略a[i],那么不能最大限度的使后续的点满足递增)
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;
    }
}