LeetCode | 0540. 有序数组中的单一元素【Python】

LeetCode 0540. Single Element in a Sorted Array有序数组中的单一元素【Medium】【Python】【二分】

Problem

LeetCode

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

力扣

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]

代码地址

GitHub链接