LeetCode - Scramble String

最新推荐文章于 2021-04-20 20:43:37 发布

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本文介绍了一种判断两个等长字符串是否可通过递归划分及子串交换的方式相互转换的算法。通过递归地将字符串划分为两个非空子串，并交换任意非叶子节点的子串顺序来实现字符串的乱序，最终判断目标字符串是否可以通过这种方式从原始字符串转化得到。

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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.

class Solution {
public:

    bool isScramble(string s1,int b1,int e1,string s2,int b2,int e2){
		string t1,t2;
		t1.assign(s1.begin()+b1,s1.begin()+e1+1);
		t2.assign(s2.begin()+b2,s2.begin()+e2+1);
		if(t1==t2){
			return true;
		}
		sort(t1.begin(),t1.end());
		sort(t2.begin(),t2.end());
		if(t1!=t2){
			return false;
		}
		if(b1==e1){
			return s1[b1]==s2[b2];
		}
		for(int i=b1;i<e1;i++){
			if(isScramble(s1,b1,i,s2,b2,b2-b1+i)&&
			isScramble(s1,i+1,e1,s2,e2-e1+i+1,e2)){
				return true;
			}
			if(isScramble(s1,b1,i,s2,e2-i+b1,e2)&&
			isScramble(s1,i+1,e1,s2,b2,e1-i-1+b2)){
				return true;
			}
		}
		return false;
    }

    bool isScramble(string s1, string s2) {
    	if(s1.size()!=s2.size()){
			return false;
		}
		return isScramble(s1,0,s1.size()-1,s2,0,s2.size()-1);
    }
};