[Leetcode] Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

 

自然语言处理（NLP）中，有一个基本问题就是求两个字符串的minimal Edit Distance, 也称Levenshtein distance。受到一篇Edit Distance介绍文章的启发，本文用动态规划求取了两个字符串之间的minimal Edit Distance. 动态规划方程将在下文进行讲解。 

1. what is minimal edit distance?

简单地说，就是仅通过插入(insert)、删除(delete)和替换(substitute)个操作将一个字符串s1变换到另一个字符串s2的最少步骤数。熟悉算法的同学很容易知道这是个动态规划问题。 

其实一个替换操作可以相当于一个delete+一个insert，所以我们将权值定义如下：

I  (insert)：1

D (delete)：1

S (substitute)：2

2. example:

intention->execution

Minimal edit distance：

delete i ; n->e ; t->x ; insert c ; n->u 求和得cost=8

3.calculate minimal edit distance dynamically
思路见注释，这里D[i,j]就是取s1前i个character和s2前j个character所得minimal edit distance

三个操作动态进行更新：

D(i,j)=min { D(i-1, j) +1, D(i, j-1) +1 , D(i-1, j-1) + s1[i]==s2[j] ? 0 : 2}；中的三项分别对应D，I，S。（详见我同学的博客）

class Solution {
public:
    int minDistance(string word1, string word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        if (len1 == 0) return len2;
        if (len2 == 0) return len1;
        vector<vector<int> > dp(len1 + 1, vector<int>(len2 + 1));
        for (int i = 0; i <= len1; ++i) dp[i][0] = i;
        for (int j = 0; j <= len2; ++j) dp[0][j] = j;
        int cost;
        for (int i = 1; i <= len1; ++i) {
            for (int j = 1; j <= len2; ++j) {
                cost = (word1[i-1] == word2[j - 1]) ? 0 : 1;
                dp[i][j] = min(dp[i-1][j-1] + cost, min(dp[i][j-1] + 1, dp[i-1][j] + 1));
            }
        }
        return dp[len1][len2];
    }
};