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.

思路1：递归的方法，简单粗暴，反复比较生成的两个子串的两种情况：

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。

注：比较的时候，保证字符串的长度是一样的否侧判断下去就没有意义了。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int nLen=s1.size();
        if(s1.compare(s2)==0)
            return true;
        string temp1(s1),temp2(s2);
        sort(temp1.begin(),temp1.end());
        sort(temp2.begin(),temp2.end());
        if(temp1.compare(temp2)!=0)
            return false;
        for(int i=1;i<nLen;i++)
        {
            string str11=s1.substr(0,i);
            string str12=s1.substr(i,nLen-i);
            string str21=s2.substr(0,i);
            string str22=s2.substr(i,nLen-i);
            if(isScramble(str11,str21)&&isScramble(str12,str22))
                return true;
            string str23=s2.substr(0,nLen-i);
            string str24=s2.substr(nLen-i,i);
            if(isScramble(str11,str24)&&isScramble(str12,str23))
                return true;
        }
        return false;
    }
};

 思路2：动态规划的思路。dp[i][j][k]表示s1从i开始k个字符和s2从j开始k个字符是否为Scramble String。

1.k==1时，dp[i][j][1]=(s1[i]==s2[j])

2.k>1时，dp[i][j][k]=(dp[i][j][index]&&dp[i+index][j+index][k-index]||dp[i][j+k-index][index]&&dp[i+index][j][k-index])，index在1~k之间。

也就是S1左==S2左 &&s1右==s2右或者s1左==s2右&&s1右==s2左。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int n=s1.size();
        if(s1.size()!=s2.size())
            return false;
        vector<vector<vector<bool>>> dp(n,vector<vector<bool> >(n,vector<bool>(n+1,false)));
        for(int k=1;k<=n;k++)
        {
            for(int i=0;i<=n-k;i++)
            {
                for(int j=0;j<=n-k;j++)
                {
                    if(k==1)
                        dp[i][j][1]=(s1[i]==s2[j]);
                    else
                    {
                        for(int index=1;index<k;index++)
                        {
                            if((dp[i][j][index]&&dp[i+index][j+index][k-index])||(dp[i][j+k-index][index]&&dp[i+index][j][k-index]))
                            {
                                dp[i][j][k]=true;
                                break;
                            }
                        }
                    }
                }
            }
        }
        return dp[0][0][n];
    }
};