[LeetCode] Scramble String（树的问题最易用递归）

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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.

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size()==0 || s2.size()==0)
            return false;
        if(s1 == s2)
            return true;
        string a1 = s1,a2 = s2;
        sort(a1.begin(),a1.end());
        sort(a2.begin(),a2.end());
        if(a1!= a2)
            return false;
        int len = s1.size();
        for(int n = 1;n < len;n++){
            if(isScramble(s1.substr(0,n),s2.substr(0,n)) && isScramble(s1.substr(n,len-n),s2.substr(n,len-n)))
                return true;
            if(isScramble(s1.substr(0,n),s2.substr(len-n,n)) && isScramble(s1.substr(n,len-n),s2.substr(0,len-n)))
                return true;
        
        }//end for
        return false;
    }//end func
};

[LeetCode] Scramble String（树的问题最易用递归）,布布扣,bubuko.com

[LeetCode] Scramble String（树的问题最易用递归）

标签：des   style   blog   color   os   io   for   ar   

原文地址：http://www.cnblogs.com/Xylophone/p/3911157.html