【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.youkuaiyun.com/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;
    }
};

转载于:https://www.cnblogs.com/ganganloveu/p/4148000.html