leetcode 87. Scramble String

87. 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.

有一点递归的感觉。

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class Solution {public:    bool iscontain(string s1,string s2)    {        sort(s1.begin(), s1.end());        sort(s2.begin(), s2.end());        return (s1 == s2);    }    bool ischeck(string s1,string s2)//第i个的前面分割，s2作为target    {        if (s1 == s2)            return 1;        if ( !iscontain(s1, s2))                  return 0;        int tsize = s2.size();        for (int i = 1; i < tsize; i++)            {                if (ischeck(s1.substr(0,i), s2.substr(0,i)) && ischeck(s1.substr(i, tsize-i),s2.substr(i, tsize-i)))                return 1;            if (ischeck(s1.substr(0,i), s2.substr(tsize-i, i)) && ischeck(s1.substr(i, tsize-i),s2.substr(0, tsize-i)))//交换                return 1;        }         return 0;    }        bool isScramble(string s1, string s2) //要有一个target    {        if (s1.size() != s2.size())            return 0;        return ischeck(s1, s2);    }};