【LeetCode】87. Scramble String

【LeetCode】87. Scramble String

【题目描述】

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.

【输入输出】

  "abcd", "bdac"                                                                                                   -> false

  "abcdefghijklmnopq", "efghijklmnopqcadb"                                                       -> false

  "oatzzffqpnwcxhejzjsnpmkmzngneo", "acegneonzmkmpnsjzjhxwnpqffzzto"    -> true

  "oad", "ado"                                                                                                       -> true

  "great", "rgtae"                                                                                                   -> true

【解题思路】

DFS+递归
1. 若s1前n项所包含的字符与s2前n项相同，递归判断s1前n项和s2前n项，s1剩下项和s2剩下项是否均为scrambled string.
2. 若s1前n项所包含的字符与s2后n项相同，递归判断s1前n项和s2后n项，s1剩下项和s2剩下项是否均为scrambled string.

【代码】

class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1 == s2) return true;
if (s1.length() != s2.length() || (s1.length() == 1 && s2.length() == 1)) return false;
bool ans = false;
for (int i = 0; i < s1.length(); i++) {
string temp1 = toSortString(s1.substr(0, i + 1));
string temp2 = toSortString(s2.substr(0, i + 1));
string temp3 = toSortString(s1.substr(s1.length() - i - 1, i + 1));
if (temp1 == temp2 && i != s1.length() - 1) ans = isScramble(s1.substr(0, i + 1), s2.substr(0, i + 1))
&& isScramble(s1.substr(i + 1, s1.length() - i - 1), s2.substr(i + 1, s1.length() - i - 1));
if (temp3 == temp2 && i != s1.length() - 1) ans = isScramble(s1.substr(0, s1.length() - i - 1), s2.substr(i + 1, s1.length() - i - 1))
&& isScramble(s1.substr(s1.length() - i - 1, i + 1), s2.substr(0, i + 1));
if (ans) return ans;
}
return false;
}

string toSortString(string s) {
string ans = s;
sort(&ans[0], &ans[0] + s.length());
return ans;
}
};