[Leetcode] Scramble String - 가장 역동적으로 기획하기 어려운 3차원

1887 단어
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.
Java:
http://blog.csdn.net/linhuanmars/article/details/24506703
s1s2 길이가 같아야 그거보다 낫죠. 사실 다 한 길이로 length 해도 되는 거 아니에요?
public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1==null||s2==null||s1.length()!=s2.length())return false;
        if(s1.length()==0) return true;
        boolean[][][] res= new boolean[s1.length()][s2.length()][s1.length()+1];
        for(int i=0;i

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