LeetCode: Scramble String

LeetCode: 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.

地址：https://oj.leetcode.com/problems/scramble-string/

算法：如果字串S1[0~i]和S2[0~i]以及S1[i+1~n]和S2[i+1~n]都满足scrambled条件，那么S1和S2也满足；如果字串S1[0~i]和S2[n-i~n]以及S1[i+1~n]和S2[0~n-i-1]满足scrambled条件，那么S1和S2也满足，其中i=1...n。注意，我们在递归调用之前，用函数isReasonString先判断一下是否有可能满足scrambled条件，这样可以减少递归的次数。代码：

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