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  • [LeetCode] Scramble String(树的问题最易用递归)

    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()==0 || s2.size()==0)
                return false;
            if(s1 == s2)
                return true;
            string a1 = s1,a2 = s2;
            sort(a1.begin(),a1.end());
            sort(a2.begin(),a2.end());
            if(a1!= a2)
                return false;
            int len = s1.size();
            for(int n = 1;n < len;n++){
                if(isScramble(s1.substr(0,n),s2.substr(0,n)) && isScramble(s1.substr(n,len-n),s2.substr(n,len-n)))
                    return true;
                if(isScramble(s1.substr(0,n),s2.substr(len-n,n)) && isScramble(s1.substr(n,len-n),s2.substr(0,len-n)))
                    return true;
            
            }//end for
            return false;
        }//end func
    };
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  • 原文地址:https://www.cnblogs.com/Xylophone/p/3911157.html
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