LeetCode 665. Non-decreasing Array

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本文探讨如何判断给定数组是否可以通过调整最多一项元素，使其成为非递减序列。提供了两种解决方案：一是限制相邻元素的升降操作，二是允许首元素自由降低，其余仅降升结合。

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
-105 <= nums[i] <= 105

---------------------------

两种方案：

方案一：降低i-1不行 AND 升高i也不行

class Solution:
    def checkPossibility(self, nums: List[int]) -> bool:
        drop_times,l = 0,len(nums)
        for i in range(1,l):
            if (nums[i] < nums[i-1]):
                drop_times += 1
                if (drop_times >= 2):
                    return False
                if ((i >= 2 and nums[i-2] > nums[i]) and (i+1 < l and nums[i-1] > nums[i+1])): #降低i-1不行 AND 升高i也不行
                    return False
        return True

方案二：除了首元素可以无限降，其他情况的下跌nums[i-1]>nums[i]，尝试把i填高，剩下的给下轮循环解决

class Solution:
    def checkPossibility(self, nums: List[int]) -> bool:
        drop_times,l = 0,len(nums)
        for i in range(1,l):
            if (nums[i] < nums[i-1]):
                drop_times += 1
                if (drop_times >= 2):
                    return False
                if (i>=2 and nums[i-2]>nums[i]):
                    nums[i] = nums[i-1]
        return True