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

一个非常经典的动态规划问题，状态转移方程：

f(i-1, j-1)                                                        word1[i] == word2[j]

f(i, j)   =

Min{ f(i-1, j), f(i, j-1), f(i-1, j-1} }  + 1          word1[i] != word2[j]

如上所示，f(i, j)表示word1.substring(i)与word2.substring(j)直接的距离

1）显然，如果word1[i] == word2[j]，转化为递归解，即f(i-1, j-1)

2）如果word1[i] == word2[j]，则可以对两个字符的最后一个字符串进行三种操作，对与字符word1[i]可以进行如下三种操作：

i) delete，即把此字符删除，则问题转化为f(i-1, j)

ii) replace，即把此字符替换为word2[j]，则问题转化为f(i-1, j-1)

iii) Insert，即在此处增加一个字符word2[j]， 则问题转化为f(i, j-1)

根据上述状态转移方程，很容易写出代码

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

又是一个比较经典的动态规划问题。

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