Scramble String

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.

题意：给出两个字符串S1,S2,问S2是不是S1的scrambled string，因为规则有点烦，这里就不加描述

题解：首先知道，字符串是被划分成两个非空的子串的，并且作为树节点的左右儿子。并且可以随意交换左右儿子，以形成新的节点，scrambled string就是这样形成的。而我们知道，其实这样的串是很多的，但是归结起来，我们可以找到规律，首先如果只是一个字符，相等就为scrambled string，反之不是。那么我们就可以用DP的思路来解决。

以DP[i][j][k]，表示S1的子串[i,i + k),     S2的子串[j,j + k)是不是互为scrambled string，定下这个状态之后，我们来划分字符串，以p为界限划分成两个子串，有(1<=p<k)，并且DP[i][j][k] = DP[i][j][p] && DP[i +p][j+p][k-p] || DP[i][j+k-p][p] && DP[i+p][j][k-p];状态转移方程就是这样了，用递归或者递推都可以做了。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1.length() != s2.length()) return false;
        strS1 = s1;
        strS2 = s2;
        return isScrambleOfSubstring(0, 0, s1.length());
    }
    bool isScrambleOfSubstring(int x, int y, int len) {
        bool flag = false;
        if (len == 0) flag = true;
        else if (len == 1) flag = (strS1[x] == strS2[y]);
        else if (unSet.find(make_pair(strS1.substr(x, len), strS2.substr(y, len)))
            != unSet.end()) flag = unSet[make_pair(strS1.substr(x, len), strS2.substr(y, len))] == true;
        else {
            for (int i = 1; i < len; i++) {
                if (flag == true) break;
                flag = (isScrambleOfSubstring(x, y, i) && isScrambleOfSubstring(x + i, y + i, len - i)) ||
                    isScrambleOfSubstring(x, y + len - i, i) && isScrambleOfSubstring(x + i, y, len - i);
            }
        }
        unSet[make_pair(strS1.substr(x, len), strS2.substr(y, len))] = flag;
        return flag;
    }
private:
    string strS1, strS2;
    typedef pair<string, string> PAIR;
    map<PAIR,bool> unSet;
};