LeetCode 题解（19）: Scramble String

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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 ifs2 is a scrambled string ofs1.

题解：

递归解法，但是无法通过时间限制。递归时那一陀计算substr的操作一定要谨慎。准备琢磨一下动态规划做法。

class Solution {public:    bool isScramble(string s1, string s2) {if(s1.length() == 1)            if(s1.compare(s2) == 0)                return true;        bool result = false;        isScrambleRecursion(s1, s2, result);        return result;    }        bool isScrambleRecursion(string s1, string s2, bool& result)    {        if(s1.length() == 1)            if(!s1.compare(s2)){return true;}        for(int div = 0; div < s1.length()-1; div++)        {result = (isScrambleRecursion(s1.substr(0, div+1), s2.substr(0, div+1), result) &&                         isScrambleRecursion(s1.substr(div+1, s1.length()-div-1), s2.substr(div+1, s2.length()-div-1), result)) ||                        (isScrambleRecursion(s1.substr(0, div+1), s2.substr(s2.length()-1-div, div+1), result)                        && isScrambleRecursion(s1.substr(div+1, s1.length()-div-1), s2.substr(0, s2.length()-div-1), result));if(result)return true;        }    }};

还是递归，进行了一下“剪枝”，大大减少了不必要的比较次数，16ms通过。

class Solution {public:    bool isScramble(string s1, string s2) {if(s1.length() == 1)            if(s1.compare(s2) == 0)                return true;        bool result = false;        isScrambleRecursion(s1, s2, result);        return result;    }        bool isScrambleRecursion(string s1, string s2, bool& result)    {        if(s1.length() == 1)            if(!s1.compare(s2)){return true;}        int A[26] = {0};          for(int i = 0; i < s1.length(); i++)              ++A[s1[i]-'a'];            for(int j = 0; j < s2.length(); j++)              --A[s2[j]-'a'];            for(int k = 0; k < 26; k++)              if(A[k] != 0) return false;                    for(int div = 0; div < s1.length()-1; div++)        {result = (isScrambleRecursion(s1.substr(0, div+1), s2.substr(0, div+1), result) &&                         isScrambleRecursion(s1.substr(div+1, s1.length()-div-1), s2.substr(div+1, s2.length()-div-1), result)) ||                        (isScrambleRecursion(s1.substr(0, div+1), s2.substr(s2.length()-1-div, div+1), result)                        && isScrambleRecursion(s1.substr(div+1, s1.length()-div-1), s2.substr(0, s2.length()-div-1), result));if(result)return true;        }    }};

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