寻找发帖水王（编程之美2.3）

#include "stdafx.h"

int find(int * ID, int N)
{
	int cand;
	int nTimes=0;
	for(int i=0;i<N;i++)
	{
		if(nTimes == 0)
		{
			cand = i;
			nTimes = 1;
		}
		else
		{
			if(cand == ID[i])
				nTimes++;
			else
				nTimes--;
		}
	}
	return cand;
}

//扩展题，有三个发帖很多的ID，发帖数目都超过了1/4
int * find2(int * ID, int N)
{
	int cand[3];
	int nTimes[3]={0};
	for(int i=0;i<N;i++)
	{
		if(nTimes[0] == 0 && cand[1] != ID[i] && cand[2] != ID[i])
		{
			cand[0]=ID[i];
			nTimes[0]=1;
		}else if(nTimes[1] == 0 && cand[0] != ID[i] && cand[2] != ID[i])
		{
			cand[1]=ID[i];
			nTimes[1]=1;
		}else if(nTimes[2] == 0 && cand[0] != ID[i] && cand[1] != ID[i])
		{
			cand[2]=ID[i];
			nTimes[2]=1;
		}else if(cand[0] != ID[i] && cand[1] != ID[i] && cand[2] != ID[i] ){ 
		   nTimes[0] -= 1; 
		   nTimes[1] -= 1; 
		   nTimes[2] -= 1; 
		 }else if(cand[0] == ID[i]){ 
			 nTimes[0] += 1; 
		 }else if(cand[1] == ID[i]){ 
			 nTimes[1] += 1; 
		 }else if(cand[2] == ID[i]){ 
			 nTimes[2] += 1; 
		 }
		
	}
	return cand;
}

int _tmain(int argc, _TCHAR* argv[])
{
	int ID1[5] = {2,3,1,2,2};
	int ID2[10] = {3,3,2,2,1,3,4,4,4};

	int n = find(ID1, 5);
	int * n2 = find2(ID2,10);
	return 0;
}