Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab"
,
Return
[
["aa","b"],
["a","a","b"]
]
将字符串划分成若干子串,每个子串都是回文的
简单DFS
class Solution {
public:
bool judge(string s)
{
for(int i=0;i<s.size()/2;i++) if(s[i] != s[s.size()-1-i]) return false;
return true;
}
void dfs(string s,vector<vector<string>>& res, vector<string>& now)
{
if(s.empty())
{
res.push_back(now);
return ;
}
for(int i=1;i<=s.size();i++)
{
if(judge(s.substr(0,i)))
{
vector<string> temp=now;
temp.push_back(s.substr(0,i));
dfs(s.substr(i), res, temp);
}
}
}
vector<vector<string>> partition(string s)
{
vector<vector<string>> res;
vector<string> temp;
dfs(s,res,temp);
return res;
}
};