Leetcode 72. Edit Distance

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本文介绍了一种使用动态规划算法来解决两个字符串之间的编辑距离问题的方法。通过插入、删除和替换字符的操作，计算从一个字符串转换为另一个字符串所需的最小步骤数。文章提供了详细的算法实现过程。

Question

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

Code

 /*
     动态规划算法解决
     */
    public int minDistance(String word1, String word2) {
        if (word1.length() == 0 || word2.length() == 0) {
            return word1.length() == 0 ? word2.length() : word1.length();
        }
        int[][] weights = new int[word1.length() + 1][word2.length() + 1];

        for (int j = 0; j <= word2.length(); j++) {
            weights[0][j] = j;
        }
        for (int i = 0; i <= word1.length(); i++) {
            weights[i][0] = 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)) {
                    weights[i][j] = weights[i - 1][j - 1];
                } else {
                    weights[i][j] = Math.min(weights[i - 1][j - 1]
                            , Math.min(weights[i - 1][j], weights[i][j - 1])) + 1;
                }
            }
        }
        return weights[word1.length()][word2.length()];
    }