Day45 ＞＞ 115、不同的子序列 + 583. 两个字符串的删除操作 + 72. 编辑距离

想逃离铁厂的老铁2026-01-19 15:28

115.不同的子序列

java 复制代码

class Solution {
    public int numDistinct(String s, String t) {
        int[][] dp = new int[s.length() + 1][t.length() + 1];
        for (int i = 0; i < s.length() + 1; i++) {
            dp[i][0] = 1;
        }
        
        for (int i = 1; i < s.length() + 1; i++) {
            for (int j = 1; j < t.length() + 1; j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }else{
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        
        return dp[s.length()][t.length()];
    }
}

583. 两个字符串的删除操作

java 复制代码

class Solution {
    public int minDistance(String word1, String word2) {
        char[] char1 = word1.toCharArray();
        char[] char2 = word2.toCharArray();

        int len1 = char1.length;
        int len2 = char2.length;

        int dp[][] = new int [len1 + 1][len2 + 1];

        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                if(char1[i - 1] == char2[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 len1 + len2 - (2 * dp[len1][len2]);
    }
}

72. 编辑距离

java 复制代码

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