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.
1 public class Solution { 2 public boolean isScramble(String s1, String s2) { 3 int len1 = s1.length(); 4 int len2 = s2.length(); 5 if(len1!=len2) return false; 6 char[] c = new char[26]; 7 for(int i=0;i<len1;i++){ 8 c[s1.charAt(i)-‘a‘]++; 9 } 10 for(int i=0;i<len2;i++){ 11 c[s2.charAt(i)-‘a‘]--; 12 } 13 for(int i=0;i<26;i++){ 14 if(c[i]!=0) return false; 15 } 16 if(len1==len2 &&len1==1) return true; 17 for(int i=0;i<len1;i++){ 18 boolean res= isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i)); 19 res |=isScramble(s1.substring(0,i),s2.substring(len2-i))&&isScramble(s1.substring(i),s2.substring(0,len2-i)); 20 if(res) return true; 21 } 22 return false; 23 24 } 25 }