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 if s2 is a scrambled string of s1.

Solution:

Code:

<span style="font-size:14px;">class Solution {public:    bool isScramble(string s1, string s2) {        int length = s1.size();        bool ***dp = new bool **[length];        for (int i = 0; i < length; i++) {            dp[i] = new bool *[length];            for (int j = 0; j < length; j++) {                dp[i][j] = new bool[length];                memset(dp[i][j], false, sizeof(bool)*length);            }        }        for (int i = length-1; i >= 0; i--)            for (int j = length-1; j >= 0; j--)                for (int k = 1; k <= length-max(i, j); k++) {                    if (s1.substr(i, k) == s2.substr(j, k))                        dp[i][j][k-1] = true;                    else {                        for (int l = 1; l <= k; l++) {                            if ((dp[i][j][l-1] && dp[i+l][j+l][k-l-1]) ||                                (dp[i][j+k-l][l-1] && dp[i+l][j][k-l-1])) {                                    dp[i][j][k-1] = true;                                    break;                            }                        }                    }                }        bool result = dp[0][0][length-1];        for (int i = 0; i < length; i++) {            for (int j = 0; j < length; j++) {                delete [] dp[i][j];            }            delete [] dp[i];        }        delete [] dp;        return result;    }};</span>