題目

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".

分析

看了好幾遍，依然覺得很難的一個題目。參考大牛的解 參考連結。 方法一：遞迴，分為下面情況 字串長度為1：很明顯，兩個字串必須完全相同才可以。
字串長度為2：當s1="ab", s2只有"ab"或者"ba"才可以。
對於任意長度的字串，我們可以把字串s1分為a1,b1兩個部分，s2分為a2,b2兩個部分，滿足（(a1~a2) && (b1~b2)）或者 （(a1~b2) && (a1~b2)）

方法二：動態規劃 遞迴解法還是很好理解的，dp這個真考察水平了，反正我還要好好琢磨琢磨。。。 使用了一個三維陣列boolean result[len][len][len],其中第一維為子串的長度，第二維為s1的起始索引，第三維為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]變化得來。

程式碼

//87. Scramble String
#include <iostream>
#include <cstdlib>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
	/*方法一：遞迴判斷，把兩個字串分別分隔為兩部分*/
	bool isScramble1(string s1, string s2) {
		if (s1.empty() && s2.empty())
			return true;

		int l1 = s1.length(), l2 = s2.length();
		if (l1 != l2)
			return false;

		if (l1 == 1)
			return s1 == s2;

		string tmp1 = s1, tmp2 = s2;
		sort(tmp1.begin(), tmp1.end());
		sort(tmp2.begin(), tmp2.end());

		for (int i = 0; i < l1; ++i)
			if (tmp1[i] != tmp2[i])
				return false;

		bool ret = false;
		//以下標i為分割點，將兩個字串分割為兩部分
		for (int i = 1; i < l1 && !ret; ++i)
		{
			string s11 = s1.substr(0, i);
			string s12 = s1.substr(i,l1-i);

			string s21 = s2.substr(0, i);
			string s22 = s2.substr(i,l1-i);

			ret = isScramble1(s11, s21) && isScramble1(s12, s22);
			if (!ret)
			{
				s21 = s2.substr(0, l1 - i);
				s22 = s2.substr(l1 - i,i);
				ret = isScramble1(s11, s22) && isScramble1(s12, s21);
			}//if
		}//for

		return ret;
	}


	/*方法二：動態規劃*/
	bool isScramble(string s1, string s2)
	{
		int l1 = s1.length(), l2 = s2.length();
		if (l1 != l2)
			return false;

		if (l1 == 0)
			return true;

		//dp[k][i][j]表示s1[i,i+k]與s2[j,j+k]的關係
		vector<vector<vector<int>>> dp(l1, vector<vector<int>>(l1, vector<int>(l1, 0)));
		for (int i = 0; i < l1; ++i)
		{
			for (int j = 0; j < l1; ++j)
			{
				if (s1[i] == s2[j])
					dp[0][i][j] = 1;
			}//for
		}//for

		for (int k = 2; k <= l1; ++k)
		{
			for (int i = l1 - k; i >= 0; --i)
			{
				for (int j = l1 - k; j >= 0; --j)
				{
					bool ret = false;
					for (int m = 1; m < k && !ret; ++m)
					{
						ret = (dp[m - 1][i][j] && dp[k - m - 1][i + m][j + m]) ||
							(dp[m - 1][i][j + k - m] && dp[k - m - 1][i + m][j]);
					}//for

					dp[k - 1][i][j] = ret ? 1 : 0;
				}//for
			}//for
		}//for
		return dp[l1 - 1][0][0];
	}


};

int main()
{
	cout << Solution().isScramble1("great", "rgtae") << endl;

	system("pause");
	return 0;
}

GitHub連結