665. Non-decreasing Array

题目

Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most one element.

We define an array is non-decreasing if nums[i] <= nums[i + 1] holds for every i (0-based) such that (0 <= i <= n - 2).

Example 1:

Input: nums = [4,2,3]
Output: true
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.

Example 2:

Input: nums = [4,2,1]
Output: false
Explanation: You can't get a non-decreasing array by modify at most one element.

Constraints:

• n == nums.length
• 1 <= n <= 104
• -10^5 <= nums[i] <= 10^5

题目大意

给你一个长度为 n 的整数数组，请你判断在 最多 改变 1 个元素的情况下，该数组能否变成一个非递减数列。我们是这样定义一个非递减数列的： 对于数组中任意的 i (0 <= i <= n-2)，总满足 nums[i] <= nums[i + 1]。

解题思路

• 简单题。循环扫描数组，找到 nums[i] > nums[i+1] 这种递减组合。一旦这种组合超过 2 组，直接返回 false。找到第一组递减组合，需要手动调节一次。如果 nums[i + 1] < nums[i - 1]，就算交换 nums[i+1] 和 nums[i]，交换结束，nums[i - 1] 仍然可能大于 nums[i + 1]，不满足题意。正确的做法应该是让较小的那个数变大，即 nums[i + 1] = nums[i]。两个元素相等满足非递减的要求。

代码

package leetcode

func checkPossibility(nums []int) bool {
	count := 0
	for i := 0; i < len(nums)-1; i++ {
		if nums[i] > nums[i+1] {
			count++
			if count > 1 {
				return false
			}
			if i > 0 && nums[i+1] < nums[i-1] {
				nums[i+1] = nums[i]
			}
		}
	}
	return true
}

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