Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab"
Output:
[
["aa","b"],
["a","a","b"]
]
題目鏈接：https://leetcode-cn.com/problems/palindrome-partitioning/
思路
回溯法練習。
法一：回溯法
紙上模擬處理情況，寫一個回溯法處理過程，再翻譯成代碼即可。
class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> res;
if(s.size()<=0) return res;
if (s.size()==1 ){
vector<string> vec = {s};
res.push_back(vec);
}else{
for( int i = 1; i<=s.size(); ++i){
string head = s.substr(0,i);
if(isPal(head)){
int sub = s.size()-i;
if(i == s.size()){
vector<string> vec = {s};
res.push_back(vec);
}else{
vector<vector<string>> vecs = partition(s.substr(i,s.size()-i));
if(vecs.size()>0){
for(int j = 0; j<vecs.size(); ++j){
vecs[j].insert(vecs[j].begin(), head);
res.push_back(vecs[j]);
}
}
}
}
}
}
return res;
}
bool isPal(string s){
if (s.size()<=0) return true;
int l = 0, r = s.size()-1;
while(l<r){
if(s[l] != s[r]) return false;
--r;
++l;
}
return true;
}
};
法二：回溯法，dp優化迴文判斷
先階段練習回溯法，之後再來寫這個代碼。