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：

#include<iostream>
using namespace std;

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

    return xor;
}

int main()
{
    int arr[] = {2 , 1 , 2 , 1 , 3 , 4 , 3};
    int length = sizeof(arr)/ sizeof(int);

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

    return 0;
   

}