首页 > 代码库 > [Leetcode] Scramble String
[Leetcode] Scramble String
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 }
[Leetcode] Scramble String
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。