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剑指 Offer II 070. 排序数组中只出现一次的数字

题目描述

给定一个只包含整数的有序数组 nums ，每个元素都会出现两次，唯有一个数只会出现一次，请找出这个唯一的数字。

 

示例 1:

输入: nums = [1,1,2,3,3,4,4,8,8]
输出: 2

示例 2:

输入: nums =  [3,3,7,7,10,11,11]
输出: 10

 

 

提示:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105

 

进阶: 采用的方案可以在 O(log n) 时间复杂度和 O(1) 空间复杂度中运行吗？

 

注意：本题与主站 540 题相同：https://leetcode.cn/problems/single-element-in-a-sorted-array/

解法

方法一

Python3

class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1
            # Equals to: if (mid % 2 == 0 and nums[mid] != nums[mid + 1]) or (mid % 2 == 1 and nums[mid] != nums[mid - 1]):
            if nums[mid] != nums[mid ^ 1]:
                right = mid
            else:
                left = mid + 1
        return nums[left]

Java

class Solution {
    public int singleNonDuplicate(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            // if ((mid % 2 == 0 && nums[mid] != nums[mid + 1]) || (mid % 2 == 1 && nums[mid] !=
            // nums[mid - 1])) {
            if (nums[mid] != nums[mid ^ 1]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return nums[left];
    }
}

C++

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + right >> 1;
            if (nums[mid] != nums[mid ^ 1])
                right = mid;
            else
                left = mid + 1;
        }
        return nums[left];
    }
};

Go

func singleNonDuplicate(nums []int) int {
	left, right := 0, len(nums)-1
	for left < right {
		mid := (left + right) >> 1
		if nums[mid] != nums[mid^1] {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return nums[left]
}

TypeScript

function singleNonDuplicate(nums: number[]): number {
    let left = 0,
        right = nums.length - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[mid] != nums[mid ^ 1]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return nums[left];
}

Swift

class Solution {
    func singleNonDuplicate(_ nums: [Int]) -> Int {
        var left = 0
        var right = nums.count - 1

        while left < right {
            let mid = (left + right) / 2
            if nums[mid] != nums[mid ^ 1] {
                right = mid
            } else {
                left = mid + 1
            }
        }

        return nums[left]
    }
}