Q72 编辑距离

给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

1. 插入一个字符
2. 删除一个字符
3. 替换一个字符

示例 1:

输入: word1 = "horse", word2 = "ros"

输出: 3

解释:

horse -> rorse (将 'h' 替换为 'r')

rorse -> rose (删除 'r')

rose -> ros (删除 'e')

示例 2:

输入: word1 = "intention", word2 = "execution"

输出: 5

解释:

intention -> inention (删除 't')

inention -> enention (将 'i' 替换为 'e')

enention -> exention (将 'n' 替换为 'x')

exention -> exection (将 'n' 替换为 'c')

exection -> execution (插入 'u')

class Solution {

    public int minDistance(String word1, String word2) {

        if (word1.length() == 0 || word2.length() == 0)

            return word1.length() == 0 ? word2.length() : word1.length();

        else if (word1.equals(word2))

            return 0;

        char[] chsA = word1.toCharArray();

        char[] chsB = word2.toCharArray();

        int[][] dp = new int[chsA.length + 1][chsB.length + 1];

        for (int i = 0; i < dp.length; i++)

            dp[i][0] = i;

        for (int i = 0; i < dp[0].length; i++)

            dp[0][i] = i;

        for (int i = 1; i < dp.length; i++) {

            for (int j = 1; j < dp[0].length; j++) {

                int t1 = dp[i][j - 1] + 1;

                int t2 = dp[i - 1][j] + 1;

                int min = t1 < t2 ? t1 : t2;

                if (chsA[i - 1] == chsB[j - 1])

                    min = dp[i - 1][j - 1] < min ? dp[i - 1][j - 1] : min;

                else

                    min = dp[i - 1][j - 1] + 1 < min ? dp[i - 1][j - 1] + 1 : min;

                dp[i][j] = min;

            }

        }

        return dp[chsA.length][chsB.length];

    }

}