[leetcode] 87. Scramble String 解题报告

题目链接：https://leetcode.com/problems/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.

思路：可以用递归来做，也可以用动态规划来做。递归的思路是两个字符串各找一个分割点，例如s1 和 s2使得要么：

1. s1分割点左边的字符串和s2分割点左边的字符串相等或者可以通过交换相等，并且s1右边的子串和s2右子串同样符合此规律

2. s1的左子串和s2的右子串相等或者可以交换使得相等，并且s1的右子串和s2的左子串符合同样的规律

如果不剪枝，会超时，可以通过判断两个字符串的字母是否个数一样来剪枝。

代码如下：

class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1 == s2) return true;
if(s1.size() != s2.size() || s1.size() ==1) return false;
int len = s1.size(), num = len;
vector<int> hash(26, 0);
for(auto val: s1) hash[val-'a']++;
for(auto val: s2) if(hash[val-'a']) hash[val-'a']--, num--;
if(num != 0) return false;
for(int i = 1; i < len; 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(len-i)) &&
isScramble(s1.substr(i), s2.substr(0, len-i))) return true;
}
return false;
}
};

参考：http://fisherlei.blogspot.com/2013/01/leetcode-scramble-string.html