LeetCode刷提笔录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":

    great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t

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".

    rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t

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".

    rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a

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.

这题我不大会...上网查了一下解法原来是用三维DP。

把s1分为两部分，s11和s12，则s2也可以分为两部分s21和s22。如果s11<-->s21 && s12<-->s22，或者s11<-->s22&&s12<-->s21，那么整个string s1和s2也scramble。

用dp[i][j][len]来表示s1以i为开头，s2以j为开头且长度为len的字串是否scramble，那么把s1(i, i+len-1)从中间分成两部分，和s2(j,j+len-1)分成的两部分分别比较是否scramble。

递推公式为dp[i][j][len] |= (dp[i][j][k] && dp[i+k][j+k][len-k]) || (dp[i][j+len-k][k] && dp[i+k][j][len-k])

base case是len为1时看s[i] == s[j]

填充一个三维数组，每次填充需要让k从1到len都走一遍，时间复杂度是O(n^4)

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length())
            return false;
        int n = s1.length();
        boolean[][][] dp = new boolean[n][n][n + 1];
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                dp[i][j][1] = s1.charAt(i) == s2.charAt(j);
            }
        }
        
        for(int len = 2; len <= n; len++){
            for(int i = 0; i < n - len + 1; i++){
                for(int j = 0; j < n - len + 1; j++){
                    for(int k = 1; k < len; k++){
                        dp[i][j][len] |= (dp[i][j][k] && dp[i + k][j + k][len - k]) || (dp[i][j + len - k][k] && dp[i + k][j][len - k]);
                    }
                }
            }
        }
        return dp[0][0][n];
    }
}