LeetCode:Single Number III

寻找唯一元素

最新推荐文章于 2025-09-07 00:48:53 发布

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最新推荐文章于 2025-09-07 00:48:53 发布

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文章标签： runtime

原文链接：http://www.cnblogs.com/xiaoying1245970347/p/5224890.html

本文介绍了一种使用位操作解决寻找数组中仅出现一次的两个元素的问题。通过异或运算找到这两个元素，并利用它们二进制表示中不同的位来区分并找出这两个特定的数值。

problem:

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

The order of the result is not important. So in the above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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Solution:Bit Manipulation

 1 class Solution {
 2 public:
 3     vector<int> singleNumber(vector<int>& nums) {
 4         
 5         int AxorB=0;
 6         for(int i=0;i<nums.size();i++)
 7         {
 8             AxorB^=nums[i];
 9         }
10         
11         //取最后一个二进制位  根据区别位将vector中的数分为两个序列
12         int mask=AxorB&(~(AxorB-1));
13         int A=0,B=0;
14         
15         for(int i=0;i<nums.size();i++)
16         {
17             if(mask&nums[i])
18                 A^=nums[i];
19             else
20                 B^=nums[i];
21                 
22         }
23         return vector<int>({A,B});
24     }
25 };