Single Number and Single Number II

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

[] 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 =  , i =  ; i < length ; ++i)
        xor = xor ^ arr[i];

    return xor;
}

完整代码：

 #include<iostream>
 using namespace std;

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

     return xor;
 }

 int main()
 {
     int arr[] = { ,  ,  ,  ,  ,  , };
     int length = sizeof(arr)/ sizeof(int);

     cout<<SingleNumber(arr , length)<<endl;

     return ;

 }