LeetCode OJ: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.

算法思想:

递归求解,需要剪枝,不然TLE

假设字符串s1:abc与s2:def,依次比较a与d是否可以并且bc与ef是否可以,或者是a与ef是否可以并且bc与d是否可以,然后循环比较ab与de是否可以并且c与f是否可以,或者是ab与d是否可以并且c与ef是否可以,只要有一组成立则abc成立

剪枝:对字符串s1与s2,如果其中有字符不一样,则不需要进行比对了,直接return false

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size()!=s2.size())return false;
        if(s1==s2)return true;
        string t1=s1,t2=s2;
        sort(t1.begin(),t1.end());
        sort(t2.begin(),t2.end());
        if(t1!=t2)return false;
        for(int i=1;i<s1.size();i++){
            if((isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i)))
            ||(isScramble(s1.substr(0,i),s2.substr(s2.size()-i))&&isScramble(s1.substr(i),s2.substr(0,s2.size()-i))))
                return true;
        }
        return false;
    }
};



LeetCode OJ:Scramble String

上一篇:Ext利用TreeStore构建动态菜单


下一篇:[CF1313D] Happy New Year - 状压dp