leetcode dfs Palindrome Partitioning
阿新 • • 發佈:2017-06-15
oid -- ati ++ eve -a cto size oss
Palindrome Partitioning
Total Accepted: 21056 Total Submissions: 81036My SubmissionsGiven 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"] ]
題意:切割字符串,使每一個子串都是回文
思路:dfs
選擇一個切割點時,假設從起點到切割點的子串是回文,那麽該切割點是合法的。能夠選擇
按這個規則dfs枚舉就能夠了
復雜度:時間O(n)。空間O(log n)
vector<vector<string> >res; bool is_palindrome(const string &s, int start, int end){ while(start < end){ if(s[start] != s[end - 1]) return false; ++start; --end; } return true; } void dfs(const string &s, int cur, vector<string>& partitions){ int size = s.size(); if(cur == size){ res.push_back(partitions); } for(int end = cur + 1; end <= size; ++end){ if(is_palindrome(s, cur, end)){ partitions.push_back(s.substr(cur, end - cur)); dfs(s, end, partitions); partitions.pop_back(); } } } vector<vector<string>> partition(string s) { vector<string> tem; dfs(s, 0, tem); return res; }
leetcode dfs Palindrome Partitioning