标签：des style blog io color os ar for sp
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:
1 public class Solution { 2 public boolean isScramble(String s1, String s2) { 3 if(s1.equals(s2)) 4 return true; 5 int l1=s1.length(); 6 int l2=s2.length(); 7 if(l1!=l2) 8 return false; 9 if(l1==0) 10 return true; 11 if(l1==1) 12 return s1.equals(s2); 13 char[] c_s1=s1.toCharArray(); 14 char[] c_s2=s2.toCharArray(); 15 Arrays.sort(c_s1); 16 Arrays.sort(c_s2); 17 for(int i=0;i<c_s1.length;++i){ 18 if(c_s1[i]!=c_s2[i]) 19 return false; 20 } 21 boolean b=false; 22 for(int i=1;i<l1&&!b;++i){ 23 String s11=s1.substring(0,i); 24 String s12=s1.substring(i); 25 String s21=s2.substring(0,i); 26 String s22=s2.substring(i); 27 b=isScramble(s11, s21)&&isScramble(s12, s22); 28 if(!b){ 29 String s31=s2.substring(0, l1-i); 30 String s32=s2.substring(l1-i); 31 b=isScramble(s11, s32)&&isScramble(s12, s31); 32 } 33 } 34 return b; 35 } 36 }
标签：des style blog io color os ar for sp
原文地址：http://www.cnblogs.com/Phoebe815/p/4067293.html
