[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.

递归剪枝。

对两个字符串可以判断排序后是否相等确认是否是scrambled的，如果不相等一定返回false,剪枝后速度加快很多。

递归过程，把s1和s2分成两个部分，sa1,sb1,sa2,sb2.

返回 ((sa1~sb1)&&(sa2~sb2))||((sa1~sb2)&&(sa2~sb1)). 

(s1~s2)表示是否scrambled

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if(len1!=len2)
            return false;
        if(s1==s2)
            return true;
        int A[26] = {0};
        for(int i=0; i<len1; ++i)
            ++A[s1[i]-'a'];
        for(int i=0; i<len2; ++i)
            --A[s2[i]-'a'];
        for(int i=0; i<26; ++i){
            if(A[i]!=0)
                return false;
        } //递归剪枝，判断字符串各个字母是否相同，采用类似桶排序简化。
        for(int i=1; i<len1; ++i){
            bool ret = isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i));
            ret = ret||(isScramble(s1.substr(0,i),s2.substr(len2-i,i))&&isScramble(s1.substr(i),s2.substr(0,len2-i)));//递归解
            if(ret)
                return true;
        }
        return false;
    }
};