LeetCode 072 Edit Distance

题目描述

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

分析

维基百科：Levenshtein distance
LeetCode 072 题解

• kitten → sitten (substitution of “s” for “k”)
• sitten → sittin (substitution of “i” for “e”)
• sittin → sitting (insertion of “g” at the end).
• 如果有word1长度为0，那么返回word2的长度
• 最大返回max{word1.length, word2.lenght}

• 只有在word1和word2字符串完全相同时，返回0

  ---

  leva,b(i,j)表示：字符串a从1至i，与字符串b从1至j，转换所需最小步数

  则题目中所求的结果是： leva,b(|a|,|b|)

• |a|：字符串a的长度
• |b|：字符串b的长度

代码

伪代码

function LevenshteinDistance(char s[1..m], char t[1..n]):  // for all i and j, d[i,j] will hold the Levenshtein distance between  // the first i characters of s and the first j characters of t;  // note that d has (m+1)*(n+1) values  declare int d[0..m, 0..n]  set each element in d to zero  // source prefixes can be transformed into empty string by  // dropping all characters  for i from 1 to m:      d[i, 0] := i  // target prefixes can be reached from empty source prefix  // by inserting every character  for j from 1 to n:      d[0, j] := j  for j from 1 to n:      for i from 1 to m:          if s[i] = t[j]:            d[i, j] := d[i-1, j-1]              // no operation required          else:            d[i, j] := minimum(d[i-1, j] + 1,   // a deletion                               d[i, j-1] + 1,   // an insertion                               d[i-1, j-1] + 1) // a substitution  return d[m, n]

题目代码

    public static int minDistance(String word1, String word2) {        if (word1.equals(word2)) {            return 0;        }        if (word1.length() == 0) {            return word2.length();        }        if (word2.length() == 0) {            return word1.length();        }        char[] w1 = word1.toCharArray();        char[] w2 = word2.toCharArray();        // (m+1)*(n+1)的矩阵        int[][] d = new int[w1.length + 1][w2.length + 1];        // 初始化第0列        for (int i = 0; i <= w1.length; i++) {            d[i][0] = i;        }        // 初始化第0行        for (int i = 0; i <= w2.length; i++) {            d[0][i] = i;        }        for (int i = 1; i <= w1.length; i++) {            for (int j = 1; j <= w2.length; j++) {                if (w1[i - 1] == w2[j - 1]) {                    d[i][j] = d[i - 1][j - 1];                } else {                    d[i][j] = Math.min(d[i - 1][j] + 1,// 删除操作                            Math.min(d[i][j - 1] + 1,// 插入操作                                    d[i - 1][j - 1] + 1));// 替换操作                }            }        }        return d[w1.length][w2.length];    }