[leetcode] Scramble String 解题报告

题目链接：https://leetcode.com/problems/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.

思路：可以用递归来做，也可以用动态规划来做。递归的思路是两个字符串各找一个分割点，例如s1 和 s2使得要么：

1. s1分割点左边的字符串和s2分割点左边的字符串相等或者可以通过交换相等，并且s1右边的子串和s2右子串同样符合此规律

2. s1的左子串和s2的右子串相等或者可以交换使得相等，并且s1的右子串和s2的左子串符合同样的规律

如果不剪枝，会超时，可以通过判断两个字符串的字母是否个数一样来剪枝。

具体代码如下：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int len1 = s1.size();
        int len2 = s2.size();
        if(len1 != len2)
            return false;
        if(s1 == s2)
            return true;
        if(len1 == 1)
            return false;
            
        int num[26];
        memset(num, 0, sizeof(num));
        for(int i =0; i< len1; i++) num[s1[i] - 'a']++;
        for(int i =0; i< len2; i++) num[s2[i] - 'a']--;
        for(int i =0; i< 26; i++) 
            if(num[i] != 0) return false;
                
        for(int i =1; i< len1; i++)
        {
            bool ans = isScramble(s1.substr(0, i), s2.substr(0,i)) 
                        && isScramble(s1.substr(i, len1-i), s2.substr(i,len2 - i));
            ans = ans || (isScramble(s1.substr(0, i), s2.substr(len2 -i, i)) 
                        && isScramble(s1.substr(i, len1-i), s2.substr(0,len1 - i)));
            if (ans == true)
                return true;
        }
        return false;
    }
};

参考：http://fisherlei.blogspot.com/2013/01/leetcode-scramble-string.html