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"]
]题目链接：https://leetcode-cn.com/problems/palindrome-partitioning/
思路
回溯法练习。
法一：回溯法
纸上模拟处理情况，写一个回溯法处理过程，再翻译成代码即可。
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;
}
};代码优化
以上代码是将下层迭代结果向上返回，需在上层多做一次for循环；如果改成向下层传递结果，递归到底后直接储存最终结果，则能提高运行效率。
class Solution {
public:
vector<vector<string>> res;
vector<vector<string>> partition(string s) {
if(s.size()<=0){
return res;
}
vector<string> cur;
trace(s, cur);
return res;
}
void trace(string s, vector<string> &cur) {
for( int i = 1; i<=s.size(); ++i){
string head = s.substr(0,i);
if(isPal(head)){
cur.push_back(head);
if(i == s.size()){
res.push_back(cur);
}else{
trace(s.substr(i,s.size()-i), cur);
}
cur.pop_back();
}
}
return;
}
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优化回文判断
先阶段练习回溯法，之后再来写这个代码。
