Single Number and Single Number II

 

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

[2] Given an array of integers, every element appears three times except for one. Find that single one. (better solution is needed)

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

题目中文意思就是说给定一个整数数组，数组中所有元素都出现了两次，只有一个元素只出现了一次，找出这个只出现了一次的元素。

用异或来解决这类问题会非常简单。

代码：

int SingleNumber(int arr[] , int length)
{
    int i , xor;
    for(xor = 0 , i = 0 ; i < length ; ++i)
        xor = xor ^ arr[i];

    return xor;
}

完整代码：

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int SingleNumber(int arr[] , int length)
 5 {
 6     int i , xor;
 7     for(xor = 0 , i = 0 ; i < length ; ++i)
 8         xor = xor ^ arr[i];
 9 
10     return xor;
11 }
12 
13 int main()
14 {
15     int arr[] = {2 , 1 , 2 , 1 , 3 , 4 , 3};
16     int length = sizeof(arr)/ sizeof(int);
17 
18     cout<<SingleNumber(arr , length)<<endl;
19 
20     return 0;
21    
22 
23 }