LeetCode| Scramble String

题目：

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.

思路：

用递归可以非常简单地完成，但是递归时需要比较两种情况：

1）s1左侧与s2左侧比较，s1右侧与s2右侧比较

2）s1左侧与s2右侧比较，s1右侧与s2左侧比较

为了简化计算量，我们可以用动态规划的方法来完成。dp[l][i][j]中的l+1代表子字符串的长度，i代表s1的起始点，j代表s2的起始点。

代码：

class Solution {public:    bool isScramble(string s1, string s2) {                if(s1.size()!=s2.size())        {            return false;        }        else        {            int len = s1.size();            bool*** dp = new bool**[len];            for(int i=0;i<len;i++)            {                dp[i]=new bool*[len];                for(int j=0;j<len;j++)                {                    dp[i][j]=new bool[len];                }            }                        for(int i=0;i<len;i++)            {                for(int j=0;j<len;j++)                {                    dp[0][i][j]=(s1[i]==s2[j]);                }            }                        for(int l=2;l<=len;l++)            {                for(int i=0;i+l<=len;i++)                {                    for(int j=0;j+l<=len;j++)                    {                        dp[l-1][i][j] = false;                        for(int k=1;k<=l-1;k++)                        {                            if(dp[k-1][i][j]&&dp[l-k-1][i+k][j+k])                            {                                dp[l-1][i][j]=true;                            }                            else if(dp[k-1][i][j+l-k]&&dp[l-k-1][i+k][j])                            {                                dp[l-1][i][j]=true;                            }                        }                    }                }            }                        return dp[len-1][0][0];        }    }};