LeetCode：665. Non-decreasing Array

题目：
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].

Answer：

class Solution(object):
    def checkPossibility(self, nums):
        p = None
        for i in range(len(nums) - 1):
            if nums[i] > nums[i+1]:
                if p is not None:
                    return False
                p = i

        return (p is None or p == 0 or p == len(nums)-2 or
                nums[p-1] <= nums[p+1] or nums[p] <= nums[p+2])

思路：
1. nums[i] > nums[i+1]出现2次或以上时，p is not None，返回False；
2. p记录了i的值，所以讨论剩下以下几种情况（注意i的范围）：

1. 当i为空，返回True；
2. 当i是首位，改变首位的值就行，返回True；
3. 当i是倒数第二个，改变最后一个值就行，返回True；
4. 当i是中间某个值，如果nums[p-1] <= nums[p+1]，nums[i]改成这中间的其中一个值就行，返回True；
5. 当i是中间某个值，如果nums[p] <= nums[p+2]，nums[i + 1]改成这中间的其中一个值就行，返回True；

6. 其它情况，返回False。