ScrambleString, 爬行字符串，动态规划

问题描述：

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.

算法分析：这个问题是字符串树通过交换叶子节点变形的。而不是简单地只要字母相同就行了。有两种方法，一种是递归，另一种是动态规划。

递归法：把s1，s2分别分成两部分，判断s1的两部分和s2的两部分是否分别可以交换相等。

//递归法
	public boolean isScramble(String s1, String s2)
	{
		if(s1.length() != s2.length())
		{
			return false;
		}
		if(s1.equals(s2))
		{
			return true;
		}
		/*
		char[] c1 = s1.toCharArray();
		Arrays.sort(c1);
		char[] c2 = s2.toCharArray();
		Arrays.sort(c2);
		if(!Arrays.equals(c1, c2))
		{
			return false;
		}
	    **/
		for(int i = 1; i < s1.length(); i ++)//i从1开始，否则i=0将无限递归
		{
			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;
    }

动态规划：

这道题用二维数组来存储中间结果已经不行了，需要一个三维数组 dp[i][j][len]，表示从s1的第i个字符开始长度为len的子串，和从s2的第j个字符开始长度为len的子串，是否互为scramble。

初始化为dp[i][j][1] = s1.charAt(i) == s2.charAt(j)，即长度为1的子串是否互为scramble。

三维数组就要三层循环，最终结果为dp[0][0][s1.length()]，即从s1的第0个字符开始长度为s1.length()的子串，即s1本身和s2本身是否互为scramble。

要判断dp[i][j][len]的值，就要把s1从i开始长度为len的串分别从k=1, 2 ... len-1处切开，判断切成的两半和s2同样切成的两半是否互为scramble，只要有一种切法满足条件，那么dp[i][j][len]就为true，否则为false。

//动态规划
	public boolean isScramble2(String s1, String s2)
	{
		if(s1.equals(s2)) return true;
		if(s1.length() != s2.length()) return false;
		//dp[i][j][len]代表s1从i，s2从j开始，长为len的字符串是否为scramble
		boolean[][][] dp = new boolean[s1.length()][s2.length()][s1.length()+1];
		for(int i = 0; i < s1.length(); i ++)
		{
			for(int j = 0; j < s2.length(); j ++)
			{
				dp[i][j][1] = (s1.charAt(i) == s2.charAt(j));
			}
		}
		for(int len = 2; len < s1.length() + 1; len ++)
		{
			for(int i = 0; i < s1.length()- len + 1; i ++)
			{
				for(int j = 0; j < s2.length() - len + 1; j ++)
				{
					for(int k = 1; k < len; k ++)
					{
						dp[i][j][len] |= (dp[i][j][k] && dp[i+k][j+k][len-k])  ||
										 (dp[i][j+len-k][k] && dp[i+k][j][len-k]);
					}
				}
			}
		}
		return dp[0][0][s1.length()];
	}