87. 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.

題意&解題思路

給一個原串s1和一個長度相同的串s2，判斷s2是否是由s1經字母二叉樹交換而來。

將s1和s2分割爲相同長度的兩個子串：

s1[0,i) 和s2[0,i)以及s1[i,len)和s2[i,len) 或
s1[0,i) 和s2[len-i,len)以及s1[i,len)和s2[0,len-i)

進行遞歸比較，若長度不同則返回false，若其子串的字母出現次數不一樣同樣也返回false，一直遞歸到單個比較字符，最後返回比較結果，代碼如下：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1 == s2)return true;
        int len = s1.length();
        int count[26] = {0};
        for(int i = 0; i < len; i++){
            count[s1[i] - 'a']++;
            count[s2[i] - 'a']--;
        }
        for(int i = 0; i < 26; i++)if(count[i] != 0)return false;
        for(int i = 1; i <= len - 1; i++){
            if(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))return true;
            if(isScramble(s1.substr(0, i), s2.substr(len - i)) && isScramble(s1.substr(i), s2.substr(0, len - i)))return true;
        }
        return false;
    }
};