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 if s2 is a scrambled string of s1.

自己的思路竟然对了，自己做这题的时候感觉自己的想法肯定会超时的，没想到通过了，不过还是有些地方可以优化的，代码不够简洁。其实想法就是对S1，S2各自可以找到一个分割点，使得S11与S21，S12与S22组成元素相等,或者S11与S22，S12与S21组成元素相等,并依此递归，若都满足则是，否则则不是，要注意，分割可能会存在好几种情况，每一种都要尝试。

自己的代码：比较low，请勿吐槽..

class Solution {
public:
bool isScramble(string s1, string s2) {
int size=s1.size();
if(size==0)
return false;
if(size==1){
if(s1==s2)
return true;
else
return false;
}
if(s1==s2)
return true;
return check(s1,s2);
}
bool check(string s1,string s2){
if(s1.size()==1){
if(s1==s2)
return true;
else
return false;
}
if(s1==s2)
return true;
int pos=-1,dir=1;
string temp1,temp2,temp3,temp4;
bool result=false;
for(int i=0;i<s1.size()-1;++i){
temp1=string(s1.begin(),s1.begin()+i+1);temp2=string(s2.begin(),s2.begin()+i+1);
temp3=string(s1.begin()+i+1,s1.end());temp4=string(s2.begin()+i+1,s2.end());
sort(temp1.begin(),temp1.end());
sort(temp2.begin(),temp2.end());
sort(temp3.begin(),temp3.end());
sort(temp4.begin(),temp4.end());
if(temp1==temp2 && temp3==temp4){
pos=i;
dir=1;
temp1=string(s1.begin(),s1.begin()+pos+1);temp2=string(s2.begin(),s2.begin()+pos+1);
temp3=string(s1.begin()+pos+1,s1.end());temp4=string(s2.begin()+pos+1,s2.end());
result=check(temp1,temp2) && check(temp3,temp4);
if(result==true)
return true;

}
temp1=string(s1.begin(),s1.begin()+i+1);temp2=string(s2.end()-i-1,s2.end());
temp3=string(s1.begin()+i+1,s1.end());temp4=string(s2.begin(),s2.end()-i-1);
sort(temp1.begin(),temp1.end());
sort(temp2.begin(),temp2.end());
sort(temp3.begin(),temp3.end());
sort(temp4.begin(),temp4.end());
if(temp1==temp2 && temp3==temp4){
pos=i;
dir=0;
temp1=string(s1.begin(),s1.begin()+pos+1);temp2=string(s2.end()-pos-1,s2.end());
temp3=string(s1.begin()+pos+1,s1.end());temp4=string(s2.begin(),s2.end()-pos-1);
result=check(temp1,temp2) && check(temp3,temp4);
if(result==true)
return true;
}
}
return false;
}
};

网上看到的比较简洁的代码：

bool isScramble(string s1, string s2) {
// Note: The Solution object is instantiated only once.
if(s1.length() != s2.length()) return false;
if(s1 == 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 i = 1; i < s1.length(); i++)
{
bool result = isScramble(s1.substr(0,i), s2.substr(0,i))
&& isScramble(s1.substr(i), s2.substr(i));
result = result || (isScramble(s1.substr(0,i), s2.substr(s2.length()-i, i))
&& isScramble(s1.substr(i), s2.substr(0,s2.length()-i)));
if(result) return true;
}
return false;
}

整体思路是一样的，但是网上这个明显更优。

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