87 - Scramble String(recursive)

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本文介绍了一种判断两个字符串是否可以通过特定规则相互转换的算法。该算法通过递归方式检查字符串能否被划分为满足条件的子串组合，适用于解决字符串乱序匹配问题。

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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.

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思路分析：

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。

/*
*/

#include "stdafx.h"
#include <iostream>
#include <algorithm>

using namespace std;

class Solution_087_ScrambleString
{
public:
	bool isScramble(string s1, string s2) 
	{
		int l1 = s1.length();
		int l2 = s2.length();

		if (l1 != l2)
		{
			return false;
		}

		if (l1 == l2)
		{
			return true;
		}

		string st1 = s1, st2 = s2;

		sort(st1.begin(), st1.end());
		sort(st2.begin(), st2.end());
		for (int i = 0; i < l1; ++i) 
		{
			if (st1[i] != st2[i]) 
			{
				return false;
			}
		}

		string s11, s12, s21, s22;
		bool res = false;

		for (int i = 1; i < l1 && !res; i++)
		{
			s11 = s1.substr(0, i);
			s12 = s1.substr(i, l1 - i);
			s21 = s2.substr(0, i);
			s22 = s2.substr(i, l1 - i);

			res = isScramble(s11, s21) && isScramble(s12, s22);

			if (!res) 
			{
				s21 = s2.substr(0, l1 - i);
				s22 = s2.substr(l1 - i, i);
				res = isScramble(s11, s22) && isScramble(s12, s21);
			}
		}

		return res;
	}
};