Leetcode 72. Edit Distance

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本文介绍了一种算法，用于计算将一个字符串(word1)转换为另一个字符串(word2)所需的最少操作数(插入、删除、替换字符)。通过动态规划方法，维护一个二维数组dp[i][j]来记录从word1的前i个字符转换到word2的前j个字符所需的最少步骤。

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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

题目大意是：有两个字符串，找到把word1 转化成 word2 最少的操作步骤。

维护一个二维数组 dp[ i ][ j ] ， dp[ i ][ j ] 表示把 word1 [0 .. i - 1] 转换到 word1 [0 .. j - 1] 所需的最小步骤。

例子： word1： abcd word2: qwer

dp[ 4 ][ 0 ] 表示把 abcd 转化成 “” 所需步骤，所以是 4.

所以

dp[ i ][ 0 ] = i

dp[ 0 ][ j ] = j

如果 word1[ i - 1] 等于 word2[ j - 1] ，dp[ i ][ j ] = dp[ i - 1][ j - 1] ，则不需要做任何改动

If they are not equal, we need to consider three cases:

Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement));
Delete word1[i - 1] and word1[0..i - 2] = word2[0..j - 1] (dp[i][j] = dp[i - 1][j] + 1 (for deletion));
Insert word2[j - 1] to word1[0..i - 1] and word1[0..i - 1] + word2[j - 1] = word2[0..j - 1] (dp[i][j] = dp[i][j - 1] + 1 (for insertion)).

替换：把word1[ i - 1] 替换成 word2[ j - 1] 则需要把在dp[ i - 1][ j - 1]的基础上加 1，所以是 dp[ i ][ j ] = dp[ i - 1][ j - 1] + 1

删除：把word1[ i - 1] 删除，使得 word1[ i - 2] 和 word2[ j - 1] 一样，需要把替换成 word2[ j - 1] , dp[ i ][ j ] = dp[ i - 1][ j ] + 1

加入：把word2[ j - 1] 加入到 word1中, dp[ i ][ j ] = dp[ i ][ j - 1] + 1

    public int minDistance(String word1, String word2) {
        int len1 = word1.length(), len2 = word2.length();        
        int[][] dp = new int[len1 + 1][len2 + 1];
        
        for (int i = 1; i <= len1; i++) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= len2; j++) {
            dp[0][j] = j;
        }
        
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = Math.min(dp[i][j - 1], Math.min(dp[i - 1][j - 1], dp[i - 1][j])) + 1;
            }
        }
        return dp[len1][len2];
    }