【LeetCode练习题】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.

解题思路：

要满足isScramble(string s1,string s2)，则必然满足isScramble(s11,s21)&&isScramble(s12,s22)或者isScramble(s11,s22)&&isScramble(s12,s21)

递归结束条件，当s1.compare(s2) == 0 时return false，即s1 和 s2 都只有一个字符且相等的时候。

当排序的sort1 ,sort2 不相等，即说明s1 和 s2 中的字符不同，return false，加上这个检查就可以大大的减少递归次数。否则就会超时。

每一次调用的s1 和 s2 的长度都是相等的，所以isScramble(s11,s21)&&isScramble(s12,s22)的时候s11.size() == s21.size()，

isScramble(s11,s22)&&isScramble(s12,s21)的时候s11.size() == s22.size()。

还有动态规划的解法，目前还不太熟，正在研究中……

代码如下：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        string sort1 = s1,sort2 = s2;
        sort(sort1.begin(),sort1.end());
        sort(sort2.begin(),sort2.end());
        if(sort1.compare(sort2) != 0) 
            return false;
        if(s1.compare(s2) == 0)
            return true;
        
        int len = s1.size();
        for(int i = 1; i < len; i++){
            string s11 = s1.substr(0,i);
            string s12 = s1.substr(i);
            string s21 = s2.substr(0,i);
            string s22 = s2.substr(i);

            if(isScramble(s11,s21)&&isScramble(s12,s22))
                return true;
            s21 = s2.substr(0,len-i);
            s22 = s2.substr(len-i);
            if(isScramble(s11,s22)&&isScramble(s12,s21))
                return true;
        }
        return false;
    }
};