583. 两个字符串的删除操作

给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。

示例：

输入: "sea", "eat"
输出: 2
解释: 第一步将"sea"变为"ea"，第二步将"eat"变为"ea"

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/delete-operation-for-two-strings
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

最长公共自序列

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            char c1 = word1.charAt(i - 1);
            for (int j = 1; j <= n; j++) {
                char c2 = word2.charAt(j - 1);
                if (c1 == c2) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        int lcs = dp[m][n];
        return m - lcs + n - lcs;
    }
}

编辑距离

import java.util.Scanner;

class Solution {
    public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        int[][] dp = new int[n + 1][m + 1];

        for (int i = 0; i <= n; ++i) {
            dp[i][0] = i;
        }

        for (int i = 0; i <= m; ++i) {
            dp[0][i] = i;
        }

        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++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], dp[i][j - 1]) + 1;
                }
            }
        }
        return dp[n][m];
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            System.out.println(new Solution().minDistance(in.next(), in.next()));
        }
    }
}