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;
}
}