-
Notifications
You must be signed in to change notification settings - Fork 2
/
Copy pathpalindromic_partioning.cpp
64 lines (40 loc) · 1.26 KB
/
palindromic_partioning.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
Given a string s, partition s such that every substring of the partition
is a palindrome. Return all possible palindrome partitioning of s.
A palindrome string is a string that reads the same backward as forward.
Example 1:
Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]
Example 2:
Input: s = "a"
Output: [["a"]]
Constraints:
1 <= s.length <= 16
s contains only lowercase English letters.
class Solution {
public:
bool isPalindrome(int start, int end, string &s) {
while (start < end) {
if (s[start++] != s[end--]) return false;
}
return true;
}
void dfs(int start, string &s, vector<string> &curr, vector<vector<string>> &res) {
if (start == s.length()) {
res.push_back(curr);
return;
}
for (int i = start; i < s.length(); i++) {
if (isPalindrome(start, i, s)) {
curr.push_back(s.substr(start, i - start + 1));
dfs(i + 1, s, curr, res);
curr.pop_back(); // backtrack
}
}
}
vector<vector<string>> partition(string s) {
vector<vector<string>> res;
vector<string> curr;
dfs(0, s, curr, res);
return res;
}
};