87. Scramble String (String; DP)

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：对付复杂问题的方法是从简单的特例来思考。简单情况：

1. 如果字符串长度为1，那么必须两个字符串完全相同；
2. 如果字符串长度为2，例如s1='ab'，则s2='ab'或s2='ba'才行
3. 如果字符串任意长度，那么可以把s1分为a1, b1两部分，s2分为a2,b2两部分。需要满足：((a1=a2)&&(b1=b2)) || ((a1=b2)&&(a2=b1)) =>可用递归

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1 == s2) return true; 
        for(int isep = 1; isep < s1.size(); ++ isep) { //traverse split pos
            string seg11 = s1.substr(0,isep);
            string seg12 = s1.substr(isep);
            
            //see if a1=a2 &&b1=b2 is ok
            string seg21 = s2.substr(0,isep);
            string seg22 = s2.substr(isep);
            if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true;
                
            //see if a1=b2 &&a2=b1 is ok
            seg21 = s2.substr(s2.size() - isep); //从后截取isep长度
            seg22 = s2.substr(0,s2.size() - isep);
            if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true;
        }
        return false;
    }
};

Result: Time Limit Exceeded

思路II: 动态规划。三维状态dp[i][j][k]，前两维分别表示s1和s2的下标起始位置，k表示子串的长度。dp[i][j][k]=true表示s1(i, i+k-1)和s2(j, j+k-1)是scramble。

状态转移方程：if(dp[i][j][split] && dp[i+split][j+split][k-split] || dp[i][j+k-split][split] && dp[i+split][j][k-split]) dp[i][j][k]=true;

因为在状态转移方程中k又要分割成更小的值，所以必须已知小值，k从小到大遍历。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int len = s1.length();
        if(len==0) return true;
        if(s1 == s2) return true;

        //初始状态
        vector<vector<vector<bool>>> dp(len, vector<vector<bool>>(len, vector<bool>(len+1, false) ) );
        for (int i = 0; i < len; ++i)
        {
            for (int j = 0; j < len; ++j)
            {
                dp[i][j][1] = s1[i]==s2[j];
            }
        }
        
        //状态转移
        for(int k = 2; k <= len; k++) //从较短的子串开始分析，为了状态转方程
        {
            for(int s1Pointer = 0; s1Pointer+k-1 < len; s1Pointer++)
            {
                for(int s2Pointer = 0; s2Pointer+k-1 < len; s2Pointer++)
                {
                    for(int split = 1; split < k; split++) //levelSize长度的任意一种分割
                    {
                        if ((dp[s1Pointer][s2Pointer][split] && dp[s1Pointer+split][s2Pointer+split][k-split]) ||
                        (dp[s1Pointer][s2Pointer+k-split][split] && dp[s1Pointer+split][s2Pointer][k-split]))
                        {
                            dp[s1Pointer][s2Pointer][k] = true;
                            break;
                        };
                    }           
                }
            }
        }
        return dp[0][0][len]; 
    }
};