Edit Distance

有两个字符串s 和 u， 问最少的步数从s转换倒u。

转换的方法，

从s任意取一个substring，然后可以执行以下三种操作

Insert one letter to any end of the string.
Delete one letter from any end of the string.
Change one letter into any other one.

也就是说从s找一个substring，使得以上三种操作的步数最小，转换倒u。

发信人: chump (chump), 信区: JobHunting
标  题: Re: 刚做了一道题挺有意思
发信站: BBS 未名空间站 (Thu Mar  1 19:25:22 2012, 美东)


我觉得就是字符串的匹配问题吧。 用u匹配s的每个等长字串，看看有多少个字符不同
。考虑到u的开始和结束的特殊情况，可以在u的前后加上足够的pad. 

#include <string>#include <iostream>using namespace std;const char PAD = '$';int match(string p, string q, size_t len){        int count = 0;        for(size_t i = 0; i < len; i++){                if (p.at(i) != q.at(i)){                        count++;                }        }        return count;}size_t edit (string from, string to, string& sub){        string pad;        pad.append( to.length()-1, PAD);        from = pad+from+pad;        int count = INT_MAX;        for (size_t i = 0; i <= from.length() - to.length(); i++ ){                string str = from.substr(i, to.length());                int cur = match(str, to, to.length());                if (cur < count){                  count = cur;                  sub = str;                }        }        while(sub.at(sub.length()-1) == PAD ){          sub = sub.substr(0, sub.length()-1);        }        while(sub.at(0) == PAD ){          sub = sub.substr(1, sub.length()-1);        }        return count;}int main() {    string a ="aabgccc";    string b = "bec";    string sub;    cout << "a=" << a <<"; b=" << b << "; edit cost =" << edit(a, b, sub) <<"; substring =" << sub << endl;    return 0;}

参考：

http://codercareer.blogspot.com/2011/12/no-25-edit-distance.html

http://www.geeksforgeeks.org/archives/13178