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题目：
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.
思路：
递归，isScramble(S1[0, i], S2[0, i]) && isScramble(S1[i, N], S2[i, N])或者isScramble(S1[0, i], S2[N-i, N]) && isScramble(S1[i, N], S2[0, i])。判断时尽早返回，比如长度不一，或者两个字符串内的字符出现的频率不一样。
package recursion; import java.util.Arrays; public class ScrambleString { public boolean isScramble(String s1, String s2) { int m = s1.length(); int n = s2.length(); if (m != n) return false; char[] c1 = s1.toCharArray(); char[] c2 = s2.toCharArray(); Arrays.sort(c1); Arrays.sort(c2); if (!isEqual(c1, c2, m)) return false; if (m == 1) return true; for (int i = 1; i < m; ++i) { if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i, m), s2.substring(i, m))) return true; if (isScramble(s1.substring(0, i), s2.substring(m - i, m)) && isScramble(s1.substring(i, m), s2.substring(0, m - i))) return true; } return false; } private boolean isEqual(char[] c1, char[] c2, int n) { for (int i = 0; i < n; ++i) { if (c1[i] != c2[i]) return false; } return true; } public static void main(String[] args) { // TODO Auto-generated method stub ScrambleString s = new ScrambleString(); System.out.println(s.isScramble("rgtae", "great")); } }
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5097721.html
