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.
Solution
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
点击查看代码
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;
}
};
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