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[leecode]Scramble String

2024-07-20 16:20:34   221人阅读

Scramble String

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.

算法思路：

分治法：其实就是暴力法= =///

代码如下：

 1 public class Solution { 2     public boolean isScramble(String s1,String s2){ 3         if(s1.length() != s2.length()) return false; 4         if(s1.length() <= 1 && s1.equals(s2)) return true; 5         if(s1.length() == 2){ 6             if(s1.equals(s2) ) return true; 7             if(s1.charAt(0) == s2.charAt(1) && s1.charAt(1) == s2.charAt(0)) return true; 8             return false; 9         }10         char[] s1Array = s1.toCharArray();11         char[] s2Array = s2.toCharArray();12         Arrays.sort(s2Array);13         Arrays.sort(s1Array);14         for(int i = 0; i < s1.length(); i++){15             if(s1Array[i] != s2Array[i])16                 return false;17         }18         boolean result = false;19         for(int i = 1; i < s1.length();i++){20             String s1pre = s1.substring(0,i);21             String s1suf = s1.substring(i);22             String s2pre = s2.substring(0, i);23             String s2suf = s2.substring(i);24             result = isScramble(s1pre, s2pre) && isScramble(s1suf, s2suf);25             if(!result){26                 String s3pre = s2.substring(0, s1.length() - i);27                 String s3suf = s2.substring(s1.length() - i);28                 result = isScramble(s1pre, s3suf) && isScramble(s1suf, s3pre);29             }30             if(result) return true;31         }32         return result;33     }34     35 }

 

[leecode]Scramble String


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