LeetCode 0540. Single Element in a Sorted Array有序数组中的单一元素【Medium】【Python】【二分】
Problem
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
Note: Your solution should run in O(log n) time and O(1) space.
问题
给定一个只包含整数的有序数组，每个元素都会出现两次，唯有一个数只会出现一次，找出这个数。
示例 1:
输入: [1,1,2,3,3,4,4,8,8]
输出: 2
示例 2:
输入: [3,3,7,7,10,11,11]
输出: 10
注意: 您的方案应该在 O(log n)时间复杂度和 O(1)空间复杂度中运行。
思路
解法一:
二分查找
每次都取偶数位置的数，如果不是偶数位置，那 mid-1。
如果 nums[mid] != nums[mid + 1]，表示单个数在 mid 左边，否则在 mid 右边。
时间复杂度: O(logn)
空间复杂度: O(1)
解法二:
判断相邻元素是否相等
每隔两个位置遍历数组，如果 nums[i] != nums[i + 1]，那单个数就是 nums[i]，否则就是最后一个数。
时间复杂度: O(n/2)
空间复杂度: O(1)
Python代码
class Solution(object):
def singleNonDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
# solution one: binary search
low, high = 0, len(nums) - 1
while low < high:
mid = int((low + high) / 2) # element in list must be int
if mid % 2 == 1: # even position
mid -= 1
if nums[mid] != nums[mid + 1]: # result is on the left of mid
high = mid
else:
low = mid + 2
return nums[low]
# # solution two: adjacent elements are equal
# for i in range(0, len(nums) - 1, 2): # step = 2
# if nums[i] != nums[i + 1]:
# return nums[i]
# return nums[-1]
