# Single Number

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?

C++代码如下：

#include<iostream>
#include<map>
using namespace std;

class Solution
{
public:
int singleNumber(int A[],int n)
{
multimap<int,int> mp;
int i;
for(i=0;i<n;i++)
{
mp.insert({A[i],i});
}
auto map_it=mp.begin();
while(map_it!=mp.end())
{
auto tmp=map_it;
tmp++;
if((tmp)==mp.end())
return (map_it)->first;
if(map_it->first!=(tmp)->first)
break;
map_it++;
map_it++;
}
return map_it->first;
}
};

int main()
{
Solution s;
int arr[]={2,4,2,4,5};
cout<<s.singleNumber(arr,5)<<endl;
}

    int singleNumber(int A[],int n)
{
multiset<int> st;
int i;
for(i=0;i<n;i++)
st.insert(A[i]);
for(i=0;i<n;i++)
if(st.count(A[i])!=2)
break;
return A[i];
}

|
6月前
|

Leetcode Single Number II （面试题推荐）

22 0
|

LeetCode 136. 只出现一次的数字 Single Number
LeetCode 136. 只出现一次的数字 Single Number
48 0
|

LeetCode 260. Single Number III

76 0
|
PHP
Leetcode 之 PHP 解析 (260. Single Number III)
Leetcode 之 PHP 解析 (260. Single Number III)
81 0
|

Leetcode-Easy 136. Single Number
Leetcode-Easy 136. Single Number
96 0
LeetCode 136：只出现一次的数字 Single Number

954 0
LeetCode 260 Single Number III（只出现一次的数字 III）（*）

1224 0
|

LeetCode 136 Single Number（只出现一次的数字）

991 0
|

LeetCode 137 Single Number II（只出现一次的数字 II）（*）

1485 0
Single Number and Single Number II
[1] Given an array of integers, every element appears twice except for one. Find that single one.
744 0