编程珠玑第三章习题5——英语中的连字符问题

编程珠玑第三章习题5——英语中的连字符问题

问题：

本问题将处理一小部分用连字符连接的英语单词方面的问题。下面的规则列表描述了一些以字母c结尾的单词的有效连字符连接：et-ic al-is-tic s-tic p-tic -lyt-ic ot-ic an-tic n-tic c-tic at-ic h-nic n-ic m-ic l-lic b-lic -clic l-ic h-ic f-ic d-ic -bic a-ic -mac i-ac  应用该规则时，必须按照上述次序进行；从而导致了"ethnic"(遵循规则"h-nic")和"clinic"(不满足前一规则，但遵循"n-ic")。如果在某个函数中给定一个单词你必须返回后缀连字符，你该如何表示这样的规则呢？

解法一：

 #include <iostream>
 #include <cstring>
 using namespace std;

 void swap(char * tem_word, int length){     //将单词首尾反序
     int i, times = length / 2;
     char ch;
     for(i = 0; i < times; ++ i){
         ch = tem_word[i];
         tem_word[i] = tem_word[length - 1 - i];
         tem_word[length - 1 - i] = ch;
     }
 }
 bool func(char * word, char * dest, int length){    //dest为欲求的word的后缀的反序
     bool flag = false;
     char arr[24][10] = {    //最长的单词的长度是 9，但列数为10，原因main函数中最后一句话
         "ci-td" , "cit-si-al" , "cit-p" , "ci-tyl-" , "ci-to" , "cit-na" , "cit-n" , "cit-c" , "cit-ta" ,
         "ci-ta" , "cin-h" , "ci-n" , "ci-m" , "cil-l" , "cil-b" , "cilc-" , "ci-l" , "ci-h" , "ci-f" ,
         "ci-d" , "cib-" , "ci-a" , "cam-" , "ca-i"
     };
     char arr1[24][8] = {
         "citd" , "citsial" , "citp" , "cityl" , "cito" , "citna" , "citn" , "citc" , "citta" ,
         "cita" , "cinh" , "cin" , "ci-m" , "cill" , "cilb" , "cilc" , "cil" , "cih" , "cif" ,
         "cid" , "cib" , "cia" , "cam" , "cai"
     };
     char * temp_str = new char[length + 1];
     strcpy(temp_str, word);
     swap(temp_str, length);
     int i, len, j;
     for(i = 0; i < 24; ++ i){
         len = strlen(arr1[i]);
         if(len > length)
             continue;
         else{
             char * bk_str = new char[len + 1];
             for(j = 0; j < len; ++ j){
                 bk_str[j] = temp_str[j];
             }
             bk_str[j] = '\0';
             if(strcmp(bk_str, arr1[i]) == 0){
                 strcpy(dest, arr[i]);
                 flag = true;
                 break;
             }else
                 continue;
         }
     }
     delete temp_str;
     return flag;
 }


 int main(){
     char * word = new char[20];
     char * dest = new char[20];;
     cin >> word;
     if(func(word, dest, strlen(word))){ //如果为真说明找到
         swap(dest,strlen(dest));
         cout << dest << endl;
     }else{
         cout << "Cannot find the bk_str ! " << endl;
     }
     delete word;
     delete dest;
     return 0;
 }
 //result:
 //clinic
 // n-ic
 //
 //ethnic
 // h-nic

解法二：思路是：hash表中带hash。第一个以ic,tic等做hash。首先根据后缀找到hash_set，再把余下的字段，依次在hash_set中查找，并取出最长的。

 #include<stdio.h>
 #include<stdlib.h>
 #include<string>
 #include<iterator>
 #include<iostream>
 #include<algorithm>
 #include<hash_map>
 #include<hash_set>

 usingnamespace std;
 usingnamespace stdext;

 char *p[] = {"et-ic","al-is-tic","s-tic","p-tic","-lyt-ic","ot-ic","an-tic",
 "n-tic","c-tic","at-ic","h-nic","n-ic","m-ic","l-lic","b-lic","-clic","l-ic",
 "h-ic","f-ic","d-ic","-bic","a-ic","-mac","i-ac"};


 void build_map(hash_map<string, hash_set<string> >& dict)
 {
 constint n = sizeof(p)/sizeof(char *);
 for (int i = 0; i < n; ++i)
 {
 string line = p[i]; reverse(line.begin(), line.end());
 int pos = line.find('-');
 dict[line.substr(0, pos)].insert(line.substr(pos + 1, line.length() - pos - 1));
 }
 }

 string lookup(hash_map<string, hash_set<string> >& dict, string word)
 {
 string line = word; reverse(line.begin(), line.end());
 int pos = line.find('-');
 string ret;

 hash_map<string, hash_set<string> >::iterator iter;
 if (dict.end() != (iter = dict.find(line.substr(0, pos))))
 {
 hash_set<string> &h = iter->second;
 string temp = line.substr(pos + 1, line.length() - pos - 1);
 for (int j = 1; j <= temp.length(); ++j)
 {
 string c = temp.substr(0, j);
 if (h.find(c) != h.end() && c.length() > ret.length())
 ret = c;
 }
 }
 ret = iter->first +"-" + ret;
 reverse(ret.begin(), ret.end());
 return ret;
 }
 int main(void)
 {
 string sline;
 hash_map<string, hash_set<string> > dict;
 build_map(dict);

 while (cin >> sline)
 {
 cout << lookup(dict, sline) << endl;
 }
 return 0;
 }

有什么好的解法可以互相探讨，谢谢！