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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.
public class Solution { public boolean isScramble(String s1, String s2) { /* 首先选择递归，简单明了，对两个string进行partition，然后比较四个字符串段。但是递归的话，这个时间复杂度比较高。然后想到能否DP，但是 即使用DP的话，也要O(n^3)。还是在递归里做些剪枝，这样就可以避免冗余计算： 对于每两个要比较的partition，需要统计他们字符出现次数（字符个数和种类要一致），如果不相等返回。*/ int s1Len=s1.length(); int s2Len=s2.length(); if(s1Len!=s2Len) return false; int frequent[]=new int[26]; for(int i=0;i<s1Len;i++){ int temp=s1.charAt(i)-‘a‘; frequent[temp]++; } for(int i=0;i<s2Len;i++){ int temp=s2.charAt(i)-‘a‘; frequent[temp]--; } for(int i=0;i<frequent.length;i++){ if(frequent[i]!=0)return false; } if(s1Len==1&&s2Len==1) return true;///// /* 报错：Line 26: java.lang.StackOverflowError if(s1.equals(s2)) return true; char[] s11=s1.toCharArray(); char[] s22=s2.toCharArray(); Arrays.sort(s11); Arrays.sort(s22); if(!(String.valueOf(s11)).equals(String.valueOf(s22))) return false;*/ for(int i=0;i<s1Len;i++){ boolean temp=isScramble(s1.substring(0,i),s2.substring(0,i))&&isScramble(s1.substring(i),s2.substring(i)); if(temp) return true; temp=isScramble(s1.substring(0,i),s2.substring(s2.length()-i))&&isScramble(s1.substring(i),s2.substring(0,s2.length()-i)); if(temp) return true; } return false; } }
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原文地址：http://www.cnblogs.com/qiaomu/p/4649870.html
