Scramble String

最新推荐文章于 2021-04-30 19:30:28 发布

原创最新推荐文章于 2021-04-30 19:30:28 发布 · 406 阅读

0 ·

CC 4.0 BY-SA版权

leetCode 同时被 2 个专栏收录

39 篇文章

订阅专栏

29 篇文章

订阅专栏

本文介绍了一种用于判断字符串是否为另一个字符串乱序排列版本的算法，通过递归分割和比较来解决字符串匹配问题。

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.

const int MAX = 100 + 10;
	int dp[MAX][MAX][MAX];
class Solution {
public:
	
	string s1,s2;
    bool isScramble(string s1, string s2) {
        if(s1.size() != s2.size())return false;
		
		this->s1 = s1;
		this->s2 = s2;

		for(int i = 0; i < s1.size(); i++)
			for(int j = 0; j < s2.size(); j++)
				for(int len = 1; len <= s1.size(); len++)
					dp[i][j][len] = -1;
		
		return match(0,0,s1.size());
    }


	bool match(int idx1, int idx2, int len)
	{
		if(dp[idx1][idx2][len] != -1)return dp[idx1][idx2][len];
		if(len == 1)
		{
			dp[idx1][idx2][len] =  s1[idx1] == s2[idx2];
			return dp[idx1][idx2][len];
		}

		for(int l = 1; l < len; l++)
		{
			if((match(idx1, idx2, l) && match(idx1 + l, idx2 + l, len - l)) || 
				(match(idx1, idx2 + len - l, l) && match(idx1 + l, idx2, len - l)))
			{
				dp[idx1][idx2][len] = 1;
			
				return 1;
			}
		}
		dp[idx1][idx2][len] = 0;
		return false;


	}

};