[leetcode]131. Palindrome Partitioning
Analysis
Happy girls day—— [每天刷題并不難0.0]
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Explanation:
Solved by recursive. Go through the string s, see if s[0, i] is palindrome . If it is, recursive call DFS(i+1) to check if the rest of s is palindrome.
Implement
class Solution {
public:
vector<vector<string>> partition(string s) {
if(s.size() == 0)
return res;
vector<string> tmp;
DFS(0, s, tmp);
return res;
}
void DFS(int idx, string s, vector<string>& tmp){
if(idx == s.size()){
res.push_back(tmp);
return ;
}
for(int i=idx; i<s.size(); i++){
if(isPalindrome(s, idx, i)){
tmp.push_back(s.substr(idx, i-idx+1));
DFS(i+1, s, tmp);
tmp.pop_back();
}
}
}
private:
vector<vector<string>> res;
bool isPalindrome(string s, int start, int end){
while(start<=end){
if(s[start++] != s[end--])
return false;
}
return true;
}
};