Given a string s
, partition s
such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s
.
A palindrome string is a string that reads the same backward as forward.
将字符串分割为所有可能的回文串。 首先用 \(check\) 来判断是否为回文串。接着用回溯来进行 \(dfs\):对当前位置 \(pos\) 和下一个位置 \(i\) 之间的字符串来判断是否为回文串,接着从 \(i+1\) 处开始
class Solution { private: vector<vector<string>> ans; vector<string> res; bool check(string x, int st, int ed){ while(st<=ed){ if(x[st++]!=x[ed--])return false; } return true; } void dfs(string s, int pos, vector<string>& res){ if(pos==s.size()){ ans.push_back(res);return; } for(int i=pos;i<s.size();i++){ string tmp = s.substr(pos,i-pos+1); if(check(s, pos, i)){ res.push_back(tmp); dfs(s, i+1, res); res.pop_back(); } } } public: vector<vector<string>> partition(string s) { dfs(s, 0, res); return ans; } };