LeetCode - Single Number

最新推荐文章于 2025-08-10 09:12:36 发布

starcroce

最新推荐文章于 2025-08-10 09:12:36 发布

阅读量516

点赞数

CC 4.0 BY-SA版权

文章标签： leetcode bit operation array

本文链接：https://blog.youkuaiyun.com/starcroce/article/details/18219459

本博客介绍了一个算法，用于在数组中找到只出现一次的数字，所有其他数字都出现了两次。通过位操作（尤其是XOR运算），可以在线性时间复杂度下解决此问题，且无需额外内存。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

http://oj.leetcode.com/problems/single-number/

Solution:
Bit operation, use XOR. We know that A^A = 0, A^0 = A. Use a for loop to XOR every element in the array and get the single one.

https://github.com/starcroce/leetcode/blob/master/single_number.cpp

// 52 ms for 14 test cases
class Solution {
public:
    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int ans = 0;
        for(int i = 0; i < n; i++) {
            ans ^= A[i];  // ^ = XOR, a^a = 0, a^a^b = b
        }
        return ans;
    }
};