Leetcode: 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 ifs2 is a scrambled string ofs1.

一开始用递归法作，结果超时。用动态规划法通过。

递归法：

class Solution {public:    bool isScramble(string s1, string s2) {        int l = s1.length();        int i = 0;                if (s1 == s2) return true;        if (l == 1) {            if (s1[0] == s2[0])                 return true;            else                 return false;        }        if (l == 2) {            if ((s1[0] == s2[0] && s1[1] == s2[1]) || (s1[0] == s2[1] && s1[1] == s2[0]))                 return true;            else                return false;        }        string s11, s12, s21, s22;        for (i = 1; i <= l-1; ++i) {            s11 = s1.substr(0, i);            s12 = s1.substr(i);            s21 = s2.substr(0, i);            s22 = s2.substr(i);            if (isScramble(s11, s21) && isScramble(s12, s22))                 return true;                                        s21 = s2.substr(l-i);            s22 = s2.substr(0, l-i);            if (isScramble(s11, s21) && isScramble(s12, s22))                 return true;                    }        return false;    }};

动态规划：

class Solution {public:    #define T(x, y, z) t[(x)*l*l + (y)*l + (z)]    bool isScramble(string s1, string s2) {        const int l = s1.length();        int *t = new int[l*l*l]();        int i = 0, j = 0, k = 0, n = 0, r = 0;        for (i = 0; i < l; ++i) {            for (j = 0; j < l; ++j) {                T(0, i, j) = s1[i] == s2[j] ? 1 : 0;            }        }        for (k = 2; k <= l; ++k) {            for (i = 0; i < l-k+1; ++i) {                for (j = 0; j < l-k+1; ++j) {                    r = 0;                    for (n = 1; n < k; ++n) {                        if ((T(k-n-1, i, j) && T(n-1, i+k-n, j+k-n)) || (T(k-n-1, i, j+n) && T(n-1, i+k-n, j))) {                            r = 1;                            break;                        }                    }                    T(k-1, i, j) = r;                }            }        }        bool ret = T(l-1, 0, 0) == 1 ? true : false;        delete[] t;        return ret;    }};