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
别人的思路:
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)
因为本题的替换操作权重同样为1,故字符不相等+1即可。
代码如下:
public class Solution {
public int minDistance(String word1, String word2) {
//边界条件
if(word1.length() == 0)
return word2.length();
if(word2.length() == 0)
return word1.length();
/*
* 本题用动态规划的解法
* f[i][j]表示word1的前i个单词到word2前j个单词的最短距离
* 状态转移方程:f[i][j] =
*/
int[][] f = new int[word1.length()][word2.length()];
boolean isEquals = false;//是否已经有相等
for(int i = 0 ; i < word2.length(); i++){
//如果相等,则距离不增加
if(word1.charAt(0) == word2.charAt(i) && !isEquals){
f[0][i] = i > 0 ? f[0][i-1]:0;//不能从0开始
isEquals = true;
}else{
f[0][i] = i > 0 ? f[0][i-1]+1:1;
}
}
isEquals = false;//是否已经有相等
for(int i = 1 ; i < word1.length(); i++){
//如果相等,则距离不增加
if(word1.charAt(i) == word2.charAt(0) && !isEquals){
f[i][0] = f[i-1][0];//不能从0开始
isEquals = true;
}else{
f[i][0] = f[i-1][0]+1;
}
}
for(int i = 1; i < word1.length();i++){
for(int j = 1; j < word2.length(); j++){
if(word1.charAt(i) == word2.charAt(j)){
f[i][j] = f[i-1][j-1];//相等的话直接相等
}else{
f[i][j] = f[i-1][j-1]+1;
}
//然后与从f[i-1][j]+1,f[i][j-1]+1比较,取最小值
f[i][j] = Math.min(f[i][j],Math.min(f[i-1][j]+1,f[i][j-1]+1));
}
}
return f[word1.length()-1][word2.length()-1];
}
}
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时间: 2024-10-23 11:18:02