## Wednesday, January 2, 2013

### [LeetCode] Scramble String 解题报告

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.
» Solve this problem

[解题思路]

• 对于每两个要比较的partition，统计他们字符出现次数，如果不相等返回。
上网搜了一下，有人说可以O(n)做出来。http://www.mitbbs.com/article_t/JobHunting/32114513.html

``````1:       bool isScramble(string s1, string s2) {
2:            // Start typing your C/C++ solution below
3:            // DO NOT write int main() function
4:            if(s1.size() != s2.size()) return false;
5:            int A[26];
6:            memset(A,0,26*sizeof(A[0]));
7:            for(int i =0;i<s1.size(); i++)
8:            {
9:                 A[s1[i]-'a']++;
10:            }
11:            for(int i =0;i<s2.size(); i++)
12:            {
13:                 A[s2[i]-'a']--;
14:            }
15:            for(int i =0;i<26; i++)
16:            {
17:                 if(A[i] !=0)
18:                 return false;
19:            }
20:            if(s1.size() ==1 && s2.size() ==1) return true;
21:            for(int i =1; i< s1.size(); i++)
22:            {
23:                 bool result= isScramble(s1.substr(0, i), s2.substr(0, i))
24:                      && isScramble(s1.substr(i, s1.size()-i), s2.substr(i, s1.size()-i));
25:                 result = result || (isScramble(s1.substr(0, i), s2.substr(s2.size() - i, i))
26:                      && isScramble(s1.substr(i, s1.size()-i), s2.substr(0, s1.size()-i)));
27:                 if(result) return true;
28:            }
29:            return false;
30:       }
``````