代码随想录算法训练营第五十六天| 583. 两个字符串的删除操作 72. 编辑距离

代码随想录算法训练营第五十六天| 583. 两个字符串的删除操作 72. 编辑距离

一、力扣583. 两个字符串的删除操作

题目链接
思路：相等时不删除，不相等时，两个字符串各删除一个，比大小，删除用步骤少的。

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 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, dp[i][j-1] + 1);
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }
}

二、力扣72. 编辑距离

题目链接
思路：区间结尾不相等，有增删换，dp[i][j] = min({dp[i - 1][j - 1], dp[me - 1][j], dp[me][j - 1]}) + 1

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 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][j-1], dp[i-1][j])) + 1;
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }
}