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.
public class Solution { public boolean isScramble(String s1, String s2) { if (s1.length() != s2.length()) { return false; } if (s1.length()==1 && s2.length()==1) { return s1.charAt(0) == s2.charAt(0); } char[] arr1 = s1.toCharArray(); char[] arr2 = s2.toCharArray(); Arrays.sort(arr1); Arrays.sort(arr2); if (!new String(arr1).equals(new String(arr2))) { return false; } if (s1.equals(s2)) { return true; } for (int split = 1; split < s1.length(); split++) { String s11 = s1.substring(0, split); String s12 = s1.substring(split); String s21 = s2.substring(0, split); String s22 = s2.substring(split); if(isScramble(s11, s21) && isScramble(s12, s22)) { return true; } s21 = s2.substring(0, s2.length() - split); s22 = s2.substring(s2.length() - split); if(isScramble(s11, s22) && isScramble(s12, s21)) { return true; } } return false; } }