LeetCode

题目:

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.

思路:

递归,isScramble(S1[0, i], S2[0, i]) && isScramble(S1[i, N], S2[i, N])或者isScramble(S1[0, i], S2[N-i, N]) && isScramble(S1[i, N], S2[0, i])。判断时尽早返回,比如长度不一,或者两个字符串内的字符出现的频率不一样。

package recursion;

import java.util.Arrays;

public class ScrambleString {

    public boolean isScramble(String s1, String s2) {
        int m = s1.length();
        int n = s2.length();
        if (m != n) return false;
        char[] c1 = s1.toCharArray();
        char[] c2 = s2.toCharArray();
        Arrays.sort(c1);
        Arrays.sort(c2);
        if (!isEqual(c1, c2, m)) return false;
        if (m == 1) return true;
        for (int i = 1; i < m; ++i) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i, m), s2.substring(i, m))) return true;
            if (isScramble(s1.substring(0, i), s2.substring(m - i, m)) && isScramble(s1.substring(i, m), s2.substring(0, m - i))) return true;
        }
        return false;
    }
    
    private boolean isEqual(char[] c1, char[] c2, int n) {
        for (int i = 0; i < n; ++i) {
            if (c1[i] != c2[i]) return false;
        }
        return true;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        ScrambleString s = new ScrambleString();
        System.out.println(s.isScramble("rgtae", "great"));
    }

}
原文地址:https://www.cnblogs.com/null00/p/5097721.html