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  • 87. 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.

    Example 1:

    Input: s1 = "great", s2 = "rgeat"
    Output: true
    

    Example 2:

    Input: s1 = "abcde", s2 = "caebd"
    Output: false

    AC code:

    class Solution {
    public:
        bool isScramble(string s1, string s2) {
            if (s1 == s2) return true;
            int len = s1.length();
            vector<int> v(26, 0);
            for (int i = 0; i < len; ++i) {
                v[s1[i]-'a']++;
                v[s2[i]-'a']--;
            }
            for (int i = 0; i < 26; ++i) {
                if (v[i] != 0) return false;
            }
            for (int i = 1; i < len; ++i) {
                if (isScramble(s1.substr(0, i), s2.substr(0, i)) && 
                    isScramble(s1.substr(i), s2.substr(i))) return true;
                if (isScramble(s1.substr(0, i), s2.substr(len-i)) &&
                    isScramble(s1.substr(i), s2.substr(0, len-i))) return true;
            }
            return false;
        }
    };
    

    Runtime: 0 ms, faster than 100.00% of C++ online submissions for Scramble String.

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/9858820.html
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