leetcode[86] Scramble String

将一个单词按照这种方式分：

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".

这样就说他们是scrambled string。现在给你两个字符串怎么判断是否属于scrambled string呢

思路：

想了n久，木有什么好的idea。

然后学习了justdoit兄的博客，理解了后自己也跟着写了个递归的。

简单的说，就是s1和s2是scramble的话，那么必然存在一个在s1上的长度l1，将s1分成s11和s12两段，同样有s21和s22。
那么要么s11和s21是scramble的并且s12和s22是scramble的；
要么s11和s22是scramble的并且s12和s21是scramble的。

判断剪枝是必要的，否则过不了大数据集合。

class Solution {
public:
bool subScramble(string s1, string s2)
{
    if (s1 == s2) return true;
    int len = s1.size();
    string sort_s1 = s1, sort_s2 = s2;
    sort(sort_s1.begin(), sort_s1.end());
    sort(sort_s2.begin(), sort_s2.end());
    if (sort_s1 != sort_s2) return false; //如果字母不相等直接返回错误
    for (int i = 1; i < len; ++i)
    {
        string s1_left = s1.substr(0,i);
        string s1_right = s1.substr(i);
        string s2_left = s2.substr(0,i);
        string s2_right = s2.substr(i);

        if (subScramble(s1_left, s2_left) && subScramble(s1_right, s2_right))
            return true;

        s2_left = s2.substr(0, len - i);
        s2_right = s2.substr(len - i);

        if (subScramble(s1_left, s2_right) && subScramble(s1_right, s2_left))
            return true;
    }
    return false;
}
bool isScramble(string s1, string s2)
{
    return subScramble(s1, s2);
}
};

leetcode[86] Scramble String

原文：http://www.cnblogs.com/higerzhang/p/4111820.html