标签：i++ push substr ash bsp vector dup word desc
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"].
UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.
class Solution { public: vector<string>combine(string word, vector<string> prev) { for(int i = 0; i < prev.size(); i++) { prev[i]+=" "+word; } return prev; } vector<string> dfs(string s, unordered_set<string>&wordDicts) { if(hash.count(s))return hash[s]; vector<string>res; if(wordDicts.count(s))res.push_back(s); for(int i = 1; i < s.length(); i++) { string word = s.substr(i); if(wordDicts.count(word)) { string rem = s.substr(0,i); vector<string> prev=combine(word,dfs(rem,wordDicts)); res.insert(res.end(),prev.begin(),prev.end()); } } hash[s] = res; return res; } vector<string> wordBreak(string s, vector<string>& wordDict) { unordered_set<string>wordDicts(wordDict.begin(), wordDict.end()); return dfs(s, wordDicts); } private: unordered_map<string, vector<string>>hash; };
标签：i++ push substr ash bsp vector dup word desc
原文地址：http://www.cnblogs.com/tsunami-lj/p/7707399.html
