如何在一篇文章中查找每个单词出现的次数

Question

主要是对一篇很长的文章,统计每个单词出现的次数，请大家给个效率比较高的方法，谢谢了!
问题补充：用vc给个具体的实现代码，假设这篇文章的每个单词已经存储到数组里了，再假设这篇文章共计20000个单词。 

Answer 1

我会选择用正则匹配整个字符串。如果是数组，先排序，然后从头到尾扫一遍就可以搞定。也可以把数组每个成员放到hashtalbe里，最后看hashtalbe的大小。或者导入数据库，用select distinct

Answer 2

用一个Node型的数组临时存储从文件中读入的单词,然后生成一个排序二叉树,中序遍历输出二叉树中节点的值;
#include <stdio.h>
#include<string.h>
struct Node
{
    char a[8];
}; /*临时存储数据的节点*/
struct Tree
{
    char b[8];
    int w;
    struct Tree *left;
    struct Tree *right;
};
void main()
{
    FILE *fp;
    struct Node m[400];
    struct Tree *root,*p,*q;
char filename[40],ch;
int i,j,k,l;
root=(struct Tree *)malloc (sizeof(struct Tree));
root=NULL;
for(i=0;i<400;i++)
{
    for(j=0;j<8;j++)
    {
        m[i].a[j]=NULL;
     }
}
printf("Please input the name of file: ");
scanf("%s", filename);
fflush(stdin);
if((fp=fopen(filename, "r")) == NULL)
{
    printf("Cannot open the file.\n");
    exit(0);
}
else
{
    i=0;
    j=0;
    ch=fgetc(fp);
    while (ch!=EOF)
    {
        if(((ch<=122)&&(ch>=65))||(ch=' '))
        {
               if(i>=8)
              {
                if(ch==' ')
                {
                       i=0;
                       j++;
                       ch=fgetc(fp);
                 }
                else
                {
                    ch=fgetc(fp);
                    continue;
                 }
               }
               if(ch==' ')
               {
                   if(m[j].a[0]!=NULL)
                   {j++;}
                   i=0;
                   ch=fgetc(fp);
               }
               else
               {
                   if((ch>=97)&&(ch<=122))
                   {ch=ch-32;}
                   m[j].a[i]=ch;
                   i++;
                   ch=fgetc(fp);
               }
           }
           else
           {
                ch=fgetc(fp);
                continue;
           }
       }
    }  /*上面部分是实现文件的输入,基本上没什么问题*/
    l=j;/*保存节点数*/
    for(i=0;i<l;i++)   /*下部分是生成二叉树*/
    {        if(m[i].a[0]==NULL)
        {break;}    /*如果临时存放数据的数组的数据域为空,则二叉树生成结束*/        else
        {
           p=(struct Tree *)malloc(sizeof(struct Tree));
           p->left=NULL;
           p->right=NULL;
           p->w=1;           for(k=0;k<8;k++)
           {
               p->b[k]=m[i].a[k];
           }           if(root==NULL)
           {root=p;}           else
           {
               q=root;
               while(q!=NULL)
               {
                   if((strcmp(q->b,m[i].a))>0)
                   {
                     if((q->left)!=NULL)
                     {
                       q=q->left;
                       if((strcmp(q->b,m[i].a))==0)
                       {
                          q->w=q->w+1;
                          q=NULL;
                       }
                      }
                     else
                     {
                        q->left=p;
                        q=NULL;
                     }
                   }                   if((strcmp(q->b,m[i].a))<0)
                   {
                      if((q->right)!=NULL)
                      {
                         q=q->right;
                         if((strcmp(q->b,m[i].a))==0)
                         {
                            q->w=q->w+1;
                            q=NULL;
                         }
                      }
                      else
                      {
                         q->right=p;
                         q=NULL;
                      }
                   }                  else
                   {
                      q->w=q->w+1;
                      q=NULL;
                   }
               }
           }
        }
    }
    fclose(fp);