You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once. Find this single element that appears only once.
Example 1:
Input: [1,1,2,3,3,4,4,8,8]
Output: 2
Example 2:
Input: [3,3,7,7,10,11,11]
Output: 10
## Solution
在有序的数组中查找，我们采用二分的方法。起始位置left = 0， right = len(nums) -1. mid =(left + right) //2.假如mid 是个偶数， 那么在它前面就是个偶数个数字，如果此时它和nums[mid+1]的 值相同，该把left 移到mid+1的位置。 同样，如果mid是奇数，如果此时。nums[mid]和nums[mid -1]相等，说明该找的数字在右边。此时也把left移到mid+1.其他情况，移动right到mid -1.
class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
left = 0
right = len(nums)-1
while (left<right):
mid = (left + right) //2
if (nums[mid]>nums[mid-1]) and (nums[mid] < nums[mid +1]):
return nums[mid]
elif((mid%2==0) and (nums[mid] ==nums[mid +1])) or ((mid%2==1) and (nums[mid] ==nums[mid -1])):
left = mid +1
else:
right = mid - 1
return nums[left]
