【LeetCode】87. Scramble String

題目：

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.

提示：

這道題能夠用遞歸方法較爲簡單地解決，遞歸的形式能夠這樣：

// recursive solution, the idea is based on the binary tree struct.
        for (int i = 1; i < s1.length(); ++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(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(0, s2.length() - i))) {
                return true;
            }
        }

結合題目的意思，若是兩個單詞是scramble的話，那麼他們可能會有非葉子節點進行了逆序的操做，所以咱們須要同時判斷：

• s1的左子節點與s2的左子節點，s1的右子節點與s2的右子節點
• s1的左子節點與s2的右子節點，s1的右子節點與s2的左子節點

另外在進行遞歸前，須要對字符串是否相等，長度是否相等，是不是變位詞進行判斷。

代碼：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        // equal ?
        if (s1 == s2) {
            return true;
        }
        // if length are differrent, they can not be scramble
        if (s1.length() != s2.length()) {
            return false;
        }
        // just like anagram
        vector<int> a(256, 0);
        for (int i = 0; i < s1.length(); ++i) {
            ++a[s1[i]];
            --a[s2[i]];
        }
        for (int i = 0; i < s1.length(); ++i) {
            if (a[s1[i]] != 0) {
                return false;
            }
        }
        // recursive solution, the idea is based on the binary tree struct.
        for (int i = 1; i < s1.length(); ++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(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(0, s2.length() - i))) {
                return true;
            }
        }
        return false;
    }
};