寻找发帖“水王” - 编程之美-2.3


http://www.nowamagic.net/librarys/veda/detail/2377
《编程之美》2.3:
Tango是微软亚洲研究院的一个试验项目。研究院的员工和实习生们都很喜欢在Tango上面交流灌水。传说,Tango有一大“水王”,他不但喜欢发贴,还会回复其他ID发的每个帖子。坊间风闻该“水王”发帖数目超过了帖子总数的一半。如果你有一个当 前论坛上所有帖子(包括回帖)的列表,其中帖子作者的ID也在表中,你能快速找出这个传说中的Tango水王吗?

Actually this is classic major element algorithm.
http://blog.csdn.net/kimili1987/article/details/8037984
题目:寻找一个ID列表中,有一个ID超过了总数的一半,找出这个ID
  1. /*寻找发帖水王,算法有些类似于寻找最大和子序列*/  
  2. Type FInd(Type * ID, int N){  
  3.     Type candicate;  
  4.     int i, nTimes;  //nTimes用来记录重复次数  
  5.     for(i = nTimes = 0; i < N; i ++){  
  6.         if(nTimes == 0){  
  7.             candicate = ID[i];                   //如果发现nTimes=0,则前面的重复完全抵消,重新选择一个candicate   
  8.             nTimes = 1;  
  9.         }else {  
  10.             if(candicate == ID[i]) nTimes ++;    //如果发现相同ID,重复次数加1   
  11.             else nTimes --;                      //否则减少   
  12.         }  
  13.     }  
  14.     return candicate;  
扩展问题
随着Tango的发展,管理员发现,“超级水王”没有了。统计结果表明,有3个发帖很多的ID,他们的发帖数目都超过了帖子总数目N的1/4。你能从发帖ID列表中快速找出他们的ID吗?

“超级水王”没有了,有3个ID发帖也很多,每个人都超过帖子总数的1/4了,怎么快速找到这三个ID?
思路和“超级水王”的是一样的,只是这次需要有三个candicate[3],和三个nTimes[3]。如果找到与这三个ID相同的,则对应的加1,如果都不相同,则三个ID数量都要减1
  1. /*寻找发帖水王,算法有些类似于寻找最大和子序列*/  
  2. Type *Find(Type * ID, int N){  
  3.     Type candicate[3];  
  4.     int i, j, k, nTimes[3];  //nTimes用来记录重复次数  
  5.     for(i = 0; i < 3; i++) nTimes[i] = 0;  
  6.     for(i = 0; i < N; i ++){          
  7.         for(j = 0; j < 3; j ++){  
  8.             if(nTimes[j] == 0){  
  9.                 candicate[j] = ID[i];                   //如果发现nTimes=0,则前面的重复完全抵消,重新选择一个candicate   
  10.                 nTimes[j] = 1;  
  11.                 break; // should break here
  12.             }else {  
  13.                 if(candicate[j] == ID[i]) {  
  14.                     nTimes[j] ++;    //如果发现相同ID,重复次数加1   
  15.                     break;  
  16.                 }  
  17.             }  
  18.         }  
  19.         //与这三个ID都不同,则都要减1   
  20.         if(j == 3){  
  21.            for(k = 0; k < 3; k++) nTimes[k]--;  
  22.         }    
  23.     }  
  24.     return candicate;  
void Find(Type* ID, int N,Type candidate[3])
{
    Type ID_NULL;//定义一个不存在的ID
    int nTimes[3], i;
    nTimes[0]=nTimes[1]=nTimes[2]=0;
    candidate[0]=candidate[1]=candidate[2]=ID_NULL;
    for(i = 0; i < N; i++)
    {
        if(ID[i]==candidate[0])
        {
             nTimes[0]++;
        }
        else if(ID[i]==candidate[1])
        {
             nTimes[1]++;
        }
        else if(ID[i]==candidate[2])
        {
             nTimes[2]++;
        }
        else if(nTimes[0]==0)
        {
             nTimes[0]=1;
             candidate[0]=ID[i];
        }
        else if(nTimes[1]==0)
        {
             nTimes[1]=1;
             candidate[1]=ID[i];
        }
        else if(nTimes[2]==0)
        {
             nTimes[2]=1;
             candidate[2]=ID[i];
        }
        else
        {
             nTimes[0]--;
             nTimes[1]--;
             nTimes[2]--;
         }
    }
    return;
}
2. Extension to K elements, each appearing at least n/(k+1) times.
   Maintain k candidates and counters.
   When considering xi, if xi equals one of the cands, increment it;
   if xi different and some counter available,
    set xi as that cand, and M = 1;
   if xi different from all, and all cands full;
    decrement them all.

3. A hot application problem in Data Stream context---heavy hitters
 in routers, top k popular items, etc...

   One important attribute of this scheme is that it uses O(1)
 memory---independent of the stream size.

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