原题网址：https://leetcode-cn.com/problems/scramble-string/description/
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.
分析：
由于一个字符串有很多种二叉表示法，貌似很难判断两个字符串是否可以做这样的变换。
对付复杂问题的方法是从简单的特例来思考，从而找出规律。
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 （(a1~b2) && (a1~b2)）
如此，我们找到了解决问题的思路。首先我们尝试用递归来写。
解法一（递归）
两个字符串的相似的必备条件是含有相同的字符集。简单的做法是把两个字符串的字符排序后，然后比较是否相同。
加上这个检查就可以大大的减少递归次数。
代码如下：
import java.util.*;
import java.lang.*;
class Program
{
public static void main(String []args){
System.out.println(isScramble("great","eatrg"));//true
System.out.println(isScramble("abcde", "caebd"));//false
}
public static boolean isScramble(String s1, String s2)
{
//using the recursive
int L1 = s1.length();
int L2 = s2.length();
if (L1 != L2) return false;
if (L1 == 0) return true;
char[] c1 = s1.toCharArray();
char[] c2 = s2.toCharArray();
if(L1==1)return c1[0]==c2[0];
Arrays.sort(c1);
Arrays.sort(c2);
for (int i = 0; i < L1; i++)
{
if (c1[i] != c2[i]) return false;
//return true;
}
boolean result = false;
for (int i = 1; i <L1 && !result; ++i)
{
String s11 = s1.substring(0, i);
String s12 = s1.substring(i);
String s21 = s2.substring(0, i);
String s22 = s2.substring(i);
result = isScramble(s11, s21) && isScramble(s12, s22);
if (!result)
{
String s31 = s2.substring(0, L1 - i);
String s32 = s2.substring(L1 - i);
result = isScramble(s11, s32) && isScramble(s12, s31);
}
}
return result;
}
}
解法二（动态规划）
这里使用了一个三维数组boolean result[len][len][len],其中第一维为子串的长度，第二维为s1的起始索引，第三维为s2的起始索引。
result[k][i][j]表示s1[i...i+k]是否可以由s2[j...j+k]变化得来。
import java.util.*;
import java.lang.*;
class Program1
{
public static void main(String []args){
System.out.println(isScramble("great","eatrg"));
System.out.println(isScramble("abcde", "caebd"));
}
public static boolean isScramble(String s1, String s2)
{
//using dynamic programming
int L1 = s1.length();
int L2 = s2.length();
if (L1 != L2) return false;
if (L1 == 0) return true;
char[] c1 = s1.toCharArray();
char[] c2 = s2.toCharArray();
int len=L1;
boolean[][][] result = new boolean[len][len][len];
for(int i=0;i<len;++i){
for(int j=0;j<len;++j){
result[0][i][j] = (c1[i]==c2[j]);
}
}
for(int k=2;k<=len;++k){
for(int i=len-k;i>=0;--i){
for(int j=len-k;j>=0;--j){
boolean r = false;
for(int m=1;m<k && !r;++m){
r = (result[m-1][i][j] && result[k-m-1][i+m][j+m]) || (result[m-1][i][j+k-m] && result[k-m-1][i+m][j]);
}
result[k-1][i][j] = r;
}
}
}
return result[len-1][0][0];
}
}
参考文章：
1.http://www.blogjava.net/sandy/archive/2013/05/22/399605.html