用相同的字母

Question

匹配的话，我想匹配的两个词，然后打印出来e.g“行为”和“猫”有'一个，对它们的C”和‘T’，使他们匹配。这里是我的代码：

#include <stdio.h> 
#include <stdlib.h> 

main() 
{ 
    FILE  *fptr; 
    char  words[100], input[100], store[1000][100] 
    char  ch 
    int   i,j,k,z,b,*ptr; 

    ptr = &b; 

    fptr = fopen("d:\\words.txt","r"); 
    if (fptr == NULL) 
    { 
      printf("Could not open file"); 
      exit(1); 
    } 

    printf("Enter the scrambled word: "); 
    fflush(stdin); 
    fgets (input,sizeof(input),stdin); 

    i = 0; 
    while (fgets(words,sizeof(words),fptr) != NULL) 
    {  
     if (strlen(input) == strlen(words)) 
     { 
      strcpy(store[i],words); 
      ++i; 
     } 
    } 
    //this is where the problem is: 
    /*am trying to match the letters in two words, if they don't match then store 1 in b, 
    if b=0 then print out the word which matched with string 'input'*/ 
    for(z = 0; z < 1000; ++z) 
    { 
     b = 0; 
     for(j = 0; j < strlen(input); ++j) 
     { 
       for(k = 0; k < strlen(store[z]); ++k) 
       { 
        if(input[j] != store[z][k]) 
         *ptr = 1;   
       } 
     } 
     if(*ptr == 0) 
     {   
        printf("Word #%2d is: %s\n", z, store[z]); 
     } 
    } 



    fflush(stdin); 
    getchar(); 
} 

请真的需要帮助。对不起，如果我没有明确表示我的问题。

Answer 1

排序两个字符串中的字母，然后比较它们是做你需要什么的更简单的方法之一。 （假设你熟悉排序）

它可能不是最有效的，但我还是那句话，太担心效率通常是最好的留到你有一个有效的解决方案和性能指标之后。

如果要检测一些更有效的方法，如果两个词是字谜，请通过垫彼得森提供的链接，Optimizing very often used anagram function

Answer 2

像这样的事情也可能工作..（对不起丑陋的识别代码，非常忙别的东西）......

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <Windows.h> 

#include <string> 
#include <list> 
#include <map> 
#include <sstream> 
#include <algorithm> 

using namespace std; 

map< string, list<string> > items; 
int c = 0; 

void readFile() { 
     FILE * f = fopen("c:\\t\\words.txt", "r"); 
     fseek(f, 0L, SEEK_END); 
     int size = ftell(f); 
     fseek(f, 0L, SEEK_SET); 
     char * data = (char*)malloc(size); 
     fread(data, size, 1, f); 

     string s = string(data); 
     istringstream reader(s); 
     while(reader) { 
      string sub; 
      reader >> sub; 

      string original = sub; 
      sort(sub.begin(), sub.end()); 

      items[sub].push_back(original);   
      c++; 
     } 


     free(data); 
     fclose(f); 
} 

bool  check(const string & v) { 
    string requestStr = v; 
    sort(requestStr.begin(), requestStr.end()); 
    printf("Requested: %s [%s]\n", v.c_str(), requestStr.c_str()); 

    if (items.find(requestStr) == items.end()) { 
     printf("Not found\n"); 
     return false; 
    } 

    list<string>::iterator it = items[requestStr].begin(); 

    while (it != items[requestStr].end()) { 
     printf("Found: %s\n", (*it).c_str());  
     it++; 
    } 
} 

int main(int argc, char ** argv) { 
    long t1 = GetTickCount(); 
    readFile(); 
    printf("Read wordlist (%i): %li ms\n", c, GetTickCount() - t1); 

    string str = "holiday"; 
    t1 = GetTickCount(); 
    check(str); 
    printf("Time: %li ms\n", GetTickCount() - t1); 


    str = "tac"; 
    t1 = GetTickCount(); 
    check(str); 
    printf("Time: %li ms\n", GetTickCount() - t1); 

    str = "dfgegs"; 
    t1 = GetTickCount(); 
    check(str); 
    printf("Time: %li ms\n", GetTickCount() - t1); 

} 

结果在109000个字文件

Read wordlist (109583): 5969 ms 
Requested: holiday [adhiloy] 
Found: holiday 
Time: 0 ms 
Requested: tac [act] 
Found: act 
Found: cat 
Time: 0 ms 
Requested: dfgegs [defggs] 
Not found 
Time: 0 ms 

120000的搜索需要7188ms，所以周围每一台S 0.0599ms earch ...