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  • 【LeetCode】87. Scramble String

    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.

    递归来做,也就是s1分为s11和s12,s2分为s21和s22。

    判断isScramble(s11,s21)&&isScramble(s12,s22)或者isScramble(s12,s21)&&isScramble(s11,s22)

    base case是字符串相同

    另外在进入递归前需要剪枝,判断两个字符串是否包含相同的字母,O(n)复杂度。

    参考http://blog.csdn.net/doc_sgl/article/details/12401335

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

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  • 原文地址:https://www.cnblogs.com/ganganloveu/p/4148000.html
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