Single Number II

寻找数组中出现一次的数

最新推荐文章于 2025-09-10 21:59:39 发布

转载最新推荐文章于 2025-09-10 21:59:39 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/YQCblog/p/3970202.html

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#runtime

本文介绍了一种算法，用于在给定数组中找出仅出现一次的元素，所有其他元素均出现三次。该算法使用哈希表实现，确保线性时间复杂度且不使用额外内存。

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

Note:

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

 1 class Solution {
 2 public:
 3     int singleNumber(int A[], int n) {
 4         
 5         unordered_map<int, int> singleMap;
 6         for(int i = 0; i < n; i++)
 7         {
 8             if(singleMap.find(A[i]) != singleMap.end())
 9                 singleMap[ A[i] ]++;
10             else
11                 singleMap[ A[i] ] = 1;
12         }
13         
14         for(unordered_map<int, int>::iterator it = singleMap.begin(); it != singleMap.end(); it++)
15         {
16             if(it->second != 3)
17                 return it->first;
18         }
19     }
20 };