LeetCode之Palindrome Partitioning

/*采用深度搜索算法。写法一。*/class Solution {public:vector<vector<string>> partition(string s) {vector<vector<string> > res;if(s.empty()) return res;vector<string> path;partition(s, res, path, -1, 0);return res;}void partition(const string &s, vector<vector<string> > &res, vector<string> &path, int pre, int cur){if(cur == s.size()-1){if(is_palindrome(s, pre+1, cur)){path.push_back(s.substr(pre+1, cur-pre));res.push_back(path);path.pop_back();}return;}else{if(is_palindrome(s, pre+1, cur)){path.push_back(s.substr(pre+1, cur-pre));partition(s, res, path, cur, cur+1);path.pop_back();}partition(s, res, path, pre, cur+1);}}bool is_palindrome(const string &s, int left, int right){if(left >= right) return true;while(left < right){if(s[left] != s[right]) return false;++left;--right;}return true;}};/*写法二。*/class Solution {public:vector<vector<string>> partition(string s) {vector<vector<string> > res;if(s.empty()) return res;vector<string> path;partition(s, res, path, 0);return res;}void partition(const string &s, vector<vector<string> > &res, vector<string> &path, int cur){if(cur == s.size()){res.push_back(path);return;}else{for(int i = cur; i < s.size(); ++i){if(is_palindrome(s, cur, i)){path.push_back(s.substr(cur, i-cur+1));partition(s, res, path, i+1);path.pop_back();}}}}bool is_palindrome(const string &s, int left, int right){if(left >= right) return true;while(left < right){if(s[left] != s[right]) return false;++left;--right;}return true;}};