LeetCode —— Scramble String

链接：http://leetcode.com/onlinejudge#question_87

原题：

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：

其实如果s2是s1的Scrambled string，那么s1也是s2的Scrambled string。

而且这个定义可以递归的进行下去，s1的左右两个子树分别为s1的两个子串，

只有当这两个子串和s2的两个对应长度的子串也互为Scrambled string的时候，s1和s2才互为Scrambled string。

所以代码就很好写了，直接递归，DFS就行了。

P.S. 这里要加个剪枝，两个串要互为Scrambled string，前提是所含的字符集一样。（不加的话，会超时）。

代码：

class Solution {public:    bool isScramble(string s1, string s2) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int length = s1.size();        if (length == 0)            return true;        if (length == 1) {            if (s1 == s2)                return true;            else                return false;        }                if (!isSameCollection(s1, s2))    return false;                for (int n=1; n<length; n++) {            string pre1 = s1.substr(0, n);            string succ1 = s1.substr(n);            if (isScramble(pre1, s2.substr(0, n)) && isScramble(succ1, s2.substr(n)))                return true;            if (isScramble(pre1, s2.substr(length-n)) && isScramble(succ1, s2.substr(0, length-n)))                return true;        }        return false;    }private:    bool isSameCollection(const string &s1, const string &s2) {const int SIZE = 26;int count1[SIZE] = { 0 };int count2[SIZE] = { 0 };for (int i=0; i<s1.size(); i++) {count1[s1[i] - 'a']++;count2[s2[i] - 'a']++;}for (int i=0; i<SIZE; i++)if (count1[i] != count2[i])return false;return true;}};