程序员面试题精选100题(63)-数组中三个只出现一次的数字

// 程序员面试题精选100题(63)-数组中三个只出现一次的数字.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
using namespace std;
#define N 5
int _tmain(int argc, _TCHAR* argv[])
{
	int arr[N]={2,14,6,6,5};// 0 is wrong
	int sum=0;
	int dior=1;
	for(int i=0;i<N;i++)
	{
		sum^=arr[i];
	}
	cout<<" in debug sum is "<<sum<<endl;
	int sum1=0,sum2=0;
	while((sum&dior)==1)// for three element or, if the result is 0 means that they are not all the same, so we look for the bit witch is 0,
		dior=dior<<1;
	cout<<" in debug dior is "<<dior<<endl;
	sum1=0;sum2=0;
	for (int i=0;i<N;i++)
	{
		if ((arr[i]&dior)==0)
		{
			sum1=sum1^arr[i];
		}
		else
		{
			sum2=sum2^arr[i];
		}
	}
	cout<<" in debug sum1,2 is "<<sum1<<" "<<sum2<<" "<<endl;
	//不知道怎么判断sum1，sum2哪个是一个单独的元素，哪个是两个元素的抑或，索性直接求了，sum3，sum4用于假设sum1是两个元素的和。
	//如果假设错了，即sum1其实是一个元素，那么经过运算求得的sum3，sum4一定有一个是为0.因为初始时sum3，sum4有且只有一个赋值为上一步的两个运算的抑或值，
	//当把所有元素分配到sum3，sum4时，原来的两个抑或的元素分别也会分配到刚才的sum3，或是sum4.这样，相当于原来的数组增加了来个元素，最后的结果只有一个单独的元素，或者在sum3，或者在sum4
	int sum3=0,sum4=0;
	int sum5=0,sum6=0;
	
	int dior1=1;
	while((sum1&dior1)==0)//deal with sum1,consider sum1 is the or value of two elements,so sum2 is another element beside current two. for two element, look for the bit which is 1,
		dior1=dior1<<1;//reset dior
	if ((sum2&dior1)==0)// if this,means offset the only element, which is just sum2
	{
		sum3=sum3^sum2;
	}
	else
	{
		sum4=sum4^sum2;
	}
	for (int i=0;i<N;i++)
	{
		if ((arr[i]&dior1)==0) //here must add () 
		{
			sum3^=arr[i];
		}
		else
		{
			sum4^=arr[i];
		}
	}
	if ((sum3!=0)&&(sum4!=0))
	{
		//sum2=sum^sum3^sum4;
		cout<<sum2<<" "<<sum3<<" "<<sum4<<endl;
	}
		
	int dior2=1;
	while((sum2&dior2)==0)//deal with sum2,consider sum2 is the or value of two elements,so sum1 is another element beside current two.
		dior2=dior2<<1;
	if ((sum1&dior2)==0)// if this,means offset the only element
	{
		sum5=sum5^sum1;
	}
	else
	{
		sum6=sum6^sum1;
	}
	for (int i=0;i<N;i++)
	{
		if ((arr[i]&dior2)==0)
		{
			sum5^=arr[i];
		}
		else
		{
			sum6^=arr[i];
		}
	}
	if ((sum5!=0)&&(sum6!=0))
	{
		//sum1=sum^sum5^sum6;
		cout<<sum1<<" "<<sum5<<" "<<sum6<<endl;
	}
	
	system("pause");
	return 0;
}

我自己的算法。看给的答案也不太简单。

有零怎么办？

不是整形怎么办？