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.
摘自:http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ascramble-string%EF%BC%8C%E4%B8%89%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/
http://blog.youkuaiyun.com/doc_sgl/article/details/12401335
http://blog.sina.com.cn/s/blog_45b8132001019ofc.html
简单的说,就是s1和s2是scramble的话,那么必然存在一个在s1上的长度l1,将s1分成s11和s12两段,同样有s21和s22。
那么要么s11和s21是scramble的并且s12和s22是scramble的;
要么s11和s22是scramble的并且s12和s21是scramble的。
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先考察简单情况:
字符串长度为1:很明显,两个字符串必须完全相同才可以。
字符串长度为2:当s1="ab", s2只有"ab"或者"ba"才可以。
对于任意长度的字符串,我们可以把字符串s1分为a1,b1两个部分,s2分为a2,b2两个部分,满足((a1~a2) && (b1~b2))或者
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注意这里的单词切分,"g r e a t",中间有4个位置,任意一个位置都算一种切分方法。
public static boolean isScramble(String s1, String s2)
{
if(s1.length() != s2.length()) return false;
if(s1 .compareTo(s2)==0) return true;
int[] A=new int[26];
for(int i = 0; i < s1.length(); i++)
A[s1.charAt(i)-'a']++;
for(int j = 0; j < s2.length(); j++)
A[s2.charAt(j)-'a']--;
for(int k = 0; k < 26; k++)
if(A[k] != 0) return false;
for(int i = 1; i < s1.length(); i++)
{
if(isScramble(s1.substring(0,i), s2.substring(0,i)) && isScramble(s1.substring(i), s2.substring(i)))
return true;
if(isScramble(s1.substring(0,i), s2.substring(s2.length()-i))&& isScramble(s1.substring(i), s2.substring(0,s2.length()-i)))
return true;
}
return false;
}