665. Non-decreasing Array

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].

非递减数列

给定一个长度为 n 的整数数组，你的任务是判断在最多改变 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： 对于数组中所有的 i (1 <= i < n)，满足 array[i] <= array[i + 1]。

示例 1:

输入: [4,2,3]
输出: True
解释: 你可以通过把第一个4变成1来使得它成为一个非递减数列。

示例 2:

输入: [4,2,1]
输出: False
解释: 你不能在只改变一个元素的情况下将其变为非递减数列。

说明:  n 的范围为 [1, 10,000]。

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关键就是当两元素nums[i - 1], nums[i]递减时

nums[i - 2], nums[ i- 1], nums[i]三元素之间的大小关系和变化关系

如果nums[i] > nums[i - 2],那么让nums[i - 1] = nums[i] 即可 如[4, 6, 5] 变为[4, 5, 5]

如果nums[i] < nums[i - 2],那么让nums[i] = nums[i - 1]即可, 如[4, 5, 3] 变为[4, 5, 5]

更改多于一次直接返回False

class Solution:
    def checkPossibility(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        modified = False

        for i in range(len(nums)):
            if i > 0:
                if nums[i] < nums[i - 1]:
                    if modified:
                        return False
                    if i > 1:
                        if nums[i] < nums[i - 2]:
                            nums[i] = nums[i - 1]
                        else:
                            nums[i - 1] = nums[i]

                    else:
                        nums[i - 1] = nums[i]
                    modified = True
        return True