Palindrome Partitioning

Desicription

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

Example:

1
2
3
4
5
6
Input: "aab"
Output:
[
["aa","b"],
["a","a","b"]
]

Solution

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
class Solution {
private:
void dfs(int index, string& s, vector<string>& path, vector<vector<string>>& res) {
if(index == s.size()) {
res.push_back(path);
return ;
}
for(int i = index; s[i]; i++) {
if(isPalindrome(s, index, i)) {
path.push_back(s.substr(index, i - index + 1));
dfs(i + 1, s, path, res);
path.pop_back();
}
}
}
bool isPalindrome(string& s, int left, int right) {
while(left <= right)
if(s[left++] != s[right--])
return false;
return true;
}
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> res;
vector<string> path;
dfs(0, s, path, res);
return res;
}
};