LeetCode87 Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. （Hard）

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.

分析：

先考虑递归地解法， 对于字符串，枚举其分割位置i - 1，只有有一种分割位置满足

(isScramble(s1.substr(0, i),s2.substr(0, i)) && isScramble(s1.substr(i, s1.size() - i),s2.substr(i, s1.size() - i))

或者

(isScramble(s1.substr(0, i),s2.substr(s1.size() - i, i)) && isScramble(s1.substr(i, s1.size() - i), s2.substr(0,s1.size() - i)))

二者之一成立，则返回true，每个位置枚举后均不成立，则返回false

注：要有一些优化，否则超时过不了，比如s1 == s2 直接返回true， 将s1 s2排个序，之后不相等则直接返回false。

代码：

1 class Solution {
2 public:
3     bool isScramble(string s1, string s2) {
4         if (s1 == s2) {
5             return true;
6         }
7         if (s1.size() != s2.size()) {
8             return false;
9         }
10         if (s1.size() == 1 && s1[0] == s2[0]) {
11             return true;
12         }
13         string cmps1 = s1;
14         string cmps2 = s2;
15         sort(cmps1.begin(), cmps1.end());
16         sort(cmps2.begin(), cmps2.end());
17         if (cmps1 != cmps2) {
18             return false;
19         }
20         for (int i = 1; i < s1.size(); ++i) {
21             if ( (isScramble(s1.substr(0, i),s2.substr(0, i))
22                   && isScramble(s1.substr(i, s1.size() - i),s2.substr(i, s1.size() - i)))
23              || (isScramble(s1.substr(0, i),s2.substr(s1.size() - i, i))
24                   && isScramble(s1.substr(i, s1.size() - i), s2.substr(0,s1.size() - i))) ) {
25                 return true;
26             }
27         }
28         return false;
29     }
30 };

好像还有三维动态规划的解法，回头再学习一个...