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 1element.

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].

解：

Find index k so that nums[k] > nums[k + 1]

• Not found: true
• Found more than one: false
• Found one:
  • If at 0 or n - 2, can easily change first or last num, true
  • Otherwise check if removing nums[k] or nums[k+1] make it non-decreasing. If can, then change nums[k] or nums[k+1] can make nums non-decreasing.

    class Solution:
        def checkPossibility(self, nums: List[int]) -> bool:
            size=len(nums)
            if size<=2:
                return True;
            k=-1
            #0-n-1  2 3 3 2 4    4 2 3  // 2 3 3 2 4 移除第k+1个//-1 4 2 3移除第k个
            for i in range(size-1):
                if nums[i]>nums[i+1]:
                    if k>=0:
                        return False
                    k=i
            if k==-1 or k==0 or k==size-2:
                return True
            #移除第k个数看剩下的是否还是非递减
            else:
                return nums[k]<=nums[k+2] or nums[k-1]<=nums[k+1]