[Leetcode] 140. Word Break II 解題報告
阿新 • • 發佈:2019-01-07
題目:
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.
Return all such possible sentences.
For example, given
s
"catsanddog"
,dict =
["cat", "cats", "and", "sand", "dog"]
.
A solution is ["cats and dog", "cat sand dog"]
.
思路:
還是DFS,但是由於字串中可能存在相同的子串,所以我們在DFS搜尋的過程中還需要記憶已經找到的結果,這樣可以大大提高效率(否則過不了大資料)。在DFS的搜尋過程中,我們首先判斷其前i個字元所構成的子字串是否存在於字典中,如果是,再遞迴地搜尋從i之後的子串所能構成的結果陣列;遞迴完成之後的關鍵一步就是合成結果,也就是將前i個字元所構成的子字串和遞迴返回的結果合併起來,作為最終結果。當然在返回之前不能忘記把它放在雜湊列表中。
程式碼:
class Solution { public: vector<string> wordBreak(string s, unordered_set<string>& wordDict) { return dfs(s, wordDict); } private: vector<string> dfs(string s, unordered_set<string>& wordDict) { if(hash.count(s) > 0) { return hash[s]; } vector<string> ret; if(wordDict.count(s) > 0) { ret.push_back(s); } for(int i = 1; i < s.length(); ++i) { string first = s.substr(0, i); if(wordDict.count(first) > 0) { vector<string> ans = dfs(s.substr(i), wordDict); for(auto val : ans) { ret.push_back(first + " " + val); } } } hash[s] = ret; return ret; } unordered_map<string, vector<string>> hash; };