《编程之美》计算字符串的相似度

来自《编程之美》3.3

题目：定义了三种不同的操作把不同的字符串变成相同

1.修改一个字符；

2.增加一个字符；

3.删除一个字符；

为了把两个字符串变得相同，每进行一次上述的某个操作，距离加1，最后定义两个字符串的相似度为“距离+1”的倒数。要求求出任意两个字符串的相似度。

解法：

#include<iostream>
usingnamespace std;
//来自编程之美3.3题，书上是用递归的方式实现的，我尝试用循环的方式
double getStrSiminarity(string str1, string str2)
{
   int len1, len2;
   len1 = str1.length();
   len2 = str2.length();
   if(len1 ==0|| len2 ==0)
       return0.0;
   //let tmp1 longer than or equal to tmp2
   string tmp1,tmp2;
   if(len1>=len2){
       tmp1 = str1;
       tmp2 = str2;
   }
   else{
       tmp1 = str2;
       tmp2 = str1;
   }
   int index1 =0;
   int index2 =0;
   double distance =0;
   while(index1<len1 || index2<len2){
       if(index2 == len2){
           distance += tmp1.substr(index1).length();
           break;
       }
       if(tmp1[index1]== tmp2[index2]){
           index1++;
           index2++;
       }
       else{
           if(tmp1.length()== tmp2.length()){
               distance ++;
               index1++;
               index2++;
           }
           else{//if tmp1 is longer than tmp2,just erase this char
               tmp1 = tmp1.erase(index1,1);
               distance++;
           }
       }
   }
           return1/(distance+1);
}
int main()
{
   string str1 ="abdd";
   string str2 ="aebdd";
   string str3 ="travelling";
   string str4 ="traveling";
   string str5 ="abcdefg";
   string str6 ="abcdef";
   string str7 ="xabcda";
   string str8 ="xfdfa";
   double test1 = getStrSiminarity(str1, str2);//0.5
   double test2 = getStrSiminarity(str3, str4);//0.5
   double test3 = getStrSiminarity(str5, str6);//0.5
   double test4 = getStrSiminarity(str7, str8);//0.2
   cout<< test1 <<"
"<<test2<<"
"<< test3 <<"
"<<test4<<endl;
   return0;
}