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[Leetcode] Scramble String
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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.
Solution:
public class Solution {
public boolean isScramble(String s1, String s2) {
if(s1.equals(s2))
return true;
int l1=s1.length();
int l2=s2.length();
if(l1!=l2)
return false;
if(l1==0)
return true;
if(l1==1)
return s1.equals(s2);
char[] c_s1=s1.toCharArray();
char[] c_s2=s2.toCharArray();
Arrays.sort(c_s1);
Arrays.sort(c_s2);
for(int i=0;i<c_s1.length;++i){
if(c_s1[i]!=c_s2[i])
return false;
}
boolean b=false;
for(int i=1;i<l1&&!b;++i){
String s11=s1.substring(0,i);
String s12=s1.substring(i);
String s21=s2.substring(0,i);
String s22=s2.substring(i);
b=isScramble(s11, s21)&&isScramble(s12, s22);
if(!b){
String s31=s2.substring(0, l1-i);
String s32=s2.substring(l1-i);
b=isScramble(s11, s32)&&isScramble(s12, s31);
}
}
return b;
}
}
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20150220:
public class Solution {
public boolean isScramble(String s1, String s2) {
if(s1==null||s2==null||s1.length()!=s2.length())
return false;
if(s1.equals(s2))
return true;
char[] c1=s1.toCharArray();
char[] c2=s2.toCharArray();
Arrays.sort(c1);
Arrays.sort(c2);
for(int i=0;i<c1.length;++i){
if(c1[i]!=c2[i])
return false;
}
int N=s1.length();
for(int i=1;i<N;++i){
String s1_temp1=s1.substring(0,i);
String s1_temp2=s1.substring(i);
String s2_temp1=s2.substring(0,i);
String s2_temp2=s2.substring(i);
if(isScramble(s1_temp1, s2_temp1)&&isScramble(s1_temp2, s2_temp2))
return true;
String s2_temp3=s2.substring(0,N-i);
String s2_temp4=s2.substring(N-i);
if(isScramble(s1_temp1, s2_temp4)&&isScramble(s1_temp2, s2_temp3))
return true;
}
return false;
}
}
