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 tTo 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 tWe 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 aWe 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.
算法思路：
分治法：其实就是暴力法= =///
代码如下：
1 public class Solution { 2 public boolean isScramble(String s1,String s2){ 3 if(s1.length() != s2.length()) return false; 4 if(s1.length() <= 1 && s1.equals(s2)) return true; 5 if(s1.length() == 2){ 6 if(s1.equals(s2) ) return true; 7 if(s1.charAt(0) == s2.charAt(1) && s1.charAt(1) == s2.charAt(0)) return true; 8 return false; 9 } 10 char[] s1Array = s1.toCharArray(); 11 char[] s2Array = s2.toCharArray(); 12 Arrays.sort(s2Array); 13 Arrays.sort(s1Array); 14 for(int i = 0; i < s1.length(); i++){ 15 if(s1Array[i] != s2Array[i]) 16 return false; 17 } 18 boolean result = false; 19 for(int i = 1; i < s1.length();i++){ 20 String s1pre = s1.substring(0,i); 21 String s1suf = s1.substring(i); 22 String s2pre = s2.substring(0, i); 23 String s2suf = s2.substring(i); 24 result = isScramble(s1pre, s2pre) && isScramble(s1suf, s2suf); 25 if(!result){ 26 String s3pre = s2.substring(0, s1.length() - i); 27 String s3suf = s2.substring(s1.length() - i); 28 result = isScramble(s1pre, s3suf) && isScramble(s1suf, s3pre); 29 } 30 if(result) return true; 31 } 32 return result; 33 } 34 35 }