Leetcode: a single element in an ordered array

Ideas ：
Strict rules logn That's two points
How to search , Just look at the small ones index Whether the parity of is the same or not
If it's two points, make an initial judgment
then left < right, mid = (left + right) // 2, Then proceed left and right Update

src：

class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        # Obviously, it's two points 
        # nums Too big to get n To do 
        n = len(nums)
        left, right = 0, n - 1
        
        if n == 1:
            return nums[0]

        if nums[left] != nums[left + 1]:
            return nums[left]
        
        if nums[right] != nums[right - 1]:
            return nums[right]
        
        def findSmallIndex(x):
            if x == 0:
                return 0
            if x == n - 1:
                return n - 2
            if nums[x] == nums[x - 1]:
                return x - 1
            if nums[x] == nums[x + 1]:
                return x
        
        while left < right:
            mid = (left + right) // 2 + 1         
            if nums[mid] != nums[mid - 1] and nums[mid] != nums[mid + 1]:
                return nums[mid]
            #  Adjust the 
            left = findSmallIndex(left)
            mid = findSmallIndex(mid)
            right = findSmallIndex(right)

            # left  To  mid There is a problem 
            if (mid - left) % 2 == 1:
                right = mid - 1
            else:
                #  forehead , Just add one 
                left = mid + 1
        
        return -1

summary ：
logn It's two points