Scramble String

最新推荐文章于 2021-04-17 00:21:09 发布

原创最新推荐文章于 2021-04-17 00:21:09 发布 · 724 阅读

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本文探讨了如何通过递归划分字符串并交换子串位置来判断一个字符串是否为另一个字符串的乱序排列。通过实现一个类`Solution`，该算法能够有效识别两个长度相等的字符串之间的乱序关系。

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.

class Solution {
public:
   	bool isScramble(string s1, string s2) {
   	    
		if(s1.size()!=s2.size()  || s1.size()==0 || s2.size()==0) return false;
	 
		if(s1==s2) return true;
		
		string ss1 = s1;
		string ss2 = s2;
		sort(ss1.begin(),ss1.end());
		sort(ss2.begin(),ss2.end());
		if(ss1!=ss2) return false;
		
		for(size_t i=1;i<s1.size();++i){
		    if(isScramble(s1.substr(0,i),s2.substr(0,i))  && isScramble(s1.substr(i,s1.size()-i),s2.substr(i,s2.size()-i))) 
			   return true;
			if(isScramble(s1.substr(0,i),s2.substr(s2.size()-i,i)) && isScramble(s1.substr(i,s1.size()-i),s2.substr(0,s2.size()-i )))
			   return true;
		 

		}
		return false;
	}
	

};