【LeetCode】87. Scramble String

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.

提示：

这道题可以用递归方法较为简单地解决，递归的形式可以这样：

// recursive solution, the idea is based on the binary tree struct.
for (int i = 1; i < s1.length(); ++i) {
if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i))) {
return true;
}
if (isScramble(s1.substr(0, i), s2.substr(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(0, s2.length() - i))) {
return true;
}
}

结合题目的意思，如果两个单词是scramble的话，那么他们可能会有非叶子节点进行了逆序的操作，因此我们需要同时判断：

s1的左子节点与s2的左子节点，s1的右子节点与s2的右子节点

s1的左子节点与s2的右子节点，s1的右子节点与s2的左子节点

另外在进行递归前，需要对字符串是否相等，长度是否相等，是否是变位词进行判断。

代码：

class Solution {
public:
bool isScramble(string s1, string s2) {
// equal ?
if (s1 == s2) {
return true;
}
// if length are differrent, they can not be scramble
if (s1.length() != s2.length()) {
return false;
}
// just like anagram
vector<int> a(256, 0);
for (int i = 0; i < s1.length(); ++i) {
++a[s1[i]];
--a[s2[i]];
}
for (int i = 0; i < s1.length(); ++i) {
if (a[s1[i]] != 0) {
return false;
}
}
// recursive solution, the idea is based on the binary tree struct.
for (int i = 1; i < s1.length(); ++i) {
if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i))) {
return true;
}
if (isScramble(s1.substr(0, i), s2.substr(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(0, s2.length() - i))) {
return true;
}
}
return false;
}
};