• LeetCode


    题目:

    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.

    思路:

    递归,isScramble(S1[0, i], S2[0, i]) && isScramble(S1[i, N], S2[i, N])或者isScramble(S1[0, i], S2[N-i, N]) && isScramble(S1[i, N], S2[0, i])。判断时尽早返回,比如长度不一,或者两个字符串内的字符出现的频率不一样。

    package recursion;
    
    import java.util.Arrays;
    
    public class ScrambleString {
    
        public boolean isScramble(String s1, String s2) {
            int m = s1.length();
            int n = s2.length();
            if (m != n) return false;
            char[] c1 = s1.toCharArray();
            char[] c2 = s2.toCharArray();
            Arrays.sort(c1);
            Arrays.sort(c2);
            if (!isEqual(c1, c2, m)) return false;
            if (m == 1) return true;
            for (int i = 1; i < m; ++i) {
                if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i, m), s2.substring(i, m))) return true;
                if (isScramble(s1.substring(0, i), s2.substring(m - i, m)) && isScramble(s1.substring(i, m), s2.substring(0, m - i))) return true;
            }
            return false;
        }
        
        private boolean isEqual(char[] c1, char[] c2, int n) {
            for (int i = 0; i < n; ++i) {
                if (c1[i] != c2[i]) return false;
            }
            return true;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            ScrambleString s = new ScrambleString();
            System.out.println(s.isScramble("rgtae", "great"));
        }
    
    }
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  • 原文地址:https://www.cnblogs.com/null00/p/5097721.html
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