算法修炼Day56|● 583. 两个字符串的删除操作 ● 72. 编辑距离

题目：583. 两个字符串的删除操作 - 力扣（LeetCode）

代码实现：

class Solution {
    public int minDistance(String word1, String word2) {
        // 只允许删除操作，最长连续公共子序列的长度
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        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] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return word1.length() + word2.length() - (dp[word1.length()][word2.length()] * 2);
    }
}

题目：72. 编辑距离 - 力扣（LeetCode）

代码实现：

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