LeetCode OJ: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.

递归求解，需要剪枝，不然TLE

假设字符串s1：abc与s2：def，依次比较a与d是否可以并且bc与ef是否可以，或者是a与ef是否可以并且bc与d是否可以，然后循环比较ab与de是否可以并且c与f是否可以，或者是ab与d是否可以并且c与ef是否可以，只要有一组成立则abc成立

剪枝：对字符串s1与s2，如果其中有字符不一样，则不需要进行比对了，直接return false

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size()!=s2.size())return false;
        if(s1==s2)return true;
        string t1=s1,t2=s2;
        sort(t1.begin(),t1.end());
        sort(t2.begin(),t2.end());
        if(t1!=t2)return false;
        for(int i=1;i