LeetCode 87. Scramble String（动态规划）

最新推荐文章于 2024-10-01 18:52:56 发布

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最新推荐文章于 2024-10-01 18:52:56 发布

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分类专栏： LeetCode 文章标签： LeetCode Hard Java 动态规划

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本文深入探讨了ScrambleString算法，一种用于判断两个字符串是否通过特定规则互相转换的复杂问题。通过动态规划方法，文章详细解释了如何确定两个等长字符串s1和s2，s2是否为s1的scramble形式。该算法利用三维动态规划数组dp，通过遍历和比较子串来高效解决这一挑战。

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Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = “great”:
在这里插入图片描述
To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node “gr” and swap its two children, it produces a scrambled string “rgeat”.
在这里插入图片描述
We say that “rgeat” is a scrambled string of “great”.

Similarly, if we continue to swap the children of nodes “eat” and “at”, it produces a scrambled string “rgtae”.
在这里插入图片描述
We say that “rgtae” is a scrambled string of “great”.

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Example 1:

Input: s1 = “great”, s2 = “rgeat”
Output: true
Example 2:

Input: s1 = “abcde”, s2 = “caebd”
Output: false

题意

任意一个字符串都可以解析成一棵二叉树，现给出两个字符串s1和s2，判断是否存在s1的二叉树表示t1和s2的二叉树表示t2，使得t1和t2只相差一次左右子树置换

思路

动态规划。三维动态规划数组dp[i][j][k]表示存在s1[i:i+k]的二叉树表示t1和s2[j:j+k]的二叉树表示t2，使得t1和t2只相差一次左右子树置换
计算dp[i][j][k]时，l从1遍历到k-1，用l对s1[i:i+k]和s2[j:j+k]进行分割，判断分割后的字符串是否相等或互为置换
注：实际代码中为了减少dp数组大小，将上述所有k映射为k-1，所有l映射为l-1

代码

/*
 * @lc app=leetcode id=87 lang=java
 *
 * [87] Scramble String
 */
class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.length() != s2.length()) {
            return false;
        }
        int len = s1.length();
        boolean[][][] dp = new boolean[len][len][len];  
        // dp[i][j][k-1] == true: s1[i:i+k] is scramble of s2[j:j+k]
        for (int k=0; k<len; k++) {
            for (int i=0; i+k<len; i++) {
                for (int j=0; j+k<len; j++) {
                    if (k == 0) {
                        dp[i][j][k] = s1.charAt(i) == s2.charAt(j);
                    } else {
                        for (int l=0; l<k && !dp[i][j][k]; l++) {
                            dp[i][j][k] = dp[i][j][l] && dp[i+l+1][j+l+1][k-l-1] ||
                            dp[i][j+k-l][l] && dp[i+l+1][j][k-l-1];
                        }
                    }
                }
            }
        }
        return dp[0][0][len-1];
    }
}