代码随想录算法训练营天 第九章 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天 第九章 五十六天| 583. 两个字符串的删除操作 72. 编辑距离

583. 两个字符串的删除操作

class Solution {
    public int minDistance(String word1, String word2) {
        // 动态规划方法1，直接算  
        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++) {
            char char1 = word1.charAt(i-1);
            for (int j = 1; j <= word2.length(); j++) {
                char char2 = word2.charAt(j-1);
                if (char1 == char2) {
                    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]+1,dp[i-1][j-1]+2));
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }
}

class Solution {
    public int minDistance(String word1, String word2) {
        // 动态规划方法2，算最长子序列
        int[][] dp = new int[word1.length()+1][word2.length()+1];
        for (int i = 1; i <= word1.length(); i++) {
            char char1 = word1.charAt(i-1);
            for (int j = 1; j <= word2.length(); j++) {
                char char2 = word2.charAt(j-1);
                if (char1 == char2) {
                    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;
    }
}

72. 编辑距离

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-1],Math.min(dp[i-1][j],dp[i][j-1])) + 1;
                }

            }
        }
        return dp[word1.length()][word2.length()];
    }
}