【代码随想录训练营】【Day54休息】【Day55】第九章｜动态规划｜子序列｜392.判断子序列｜115.不同的子序列｜583.两个字符串的删除操作

判断子序列

题目详细：LeetCode.392

详细的题解可查阅：《代码随想录》— 判断子序列

Java解法（动态规划）：

class Solution {
    public boolean isSubsequence(String s, String t) {
        int n = s.length(), m = t.length();
        if(n > m) return false;
        int[][] dp = new int[n + 1][m + 1];
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(s.charAt(i - 1) == t.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 dp[n][m] == s.length();
    }
}

---

不同的子序列

题目详细：LeetCode.115

详细的题解可查阅：《代码随想录》— 不同的子序列

Java解法（动态规划）：

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

---

两个字符串的删除操作

题目详细：LeetCode.583

详细的题解可查阅：《代码随想录》— 两个字符串的删除操作

Java解法（动态规划一）：

class Solution {
    public int minDistance(String word1, String word2) {
        int n = word1.length(), 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][j - 1], dp[i - 1][j]) + 1;
                }
            }
        }
        return dp[n][m];
    }
}

Java解法（动态规划二）：

class Solution {
    public int minDistance(String word1, String word2) {
        int n = word1.length(), m = word2.length();
        int[][] dp = new int[n + 1][m + 1];
        // 求出两个字符串的最长公共子序列长度
        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] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        // 最后用两个字符串的总长度减去两个最长公共子序列的长度就是删除的最少步数
        return n + m - dp[n][m] * 2;
    }
}

---