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https://leetcode.com/problems/word-break/
Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "leetcode",
dict = ["leet", "code"].
Return true because "leetcode" can be segmented as "leet code".
class Solution { public: bool wordBreak(string s, unordered_set<string>& wordDict) { if(s.length() == 0) return false; vector<bool> canBreak(s.length(), false); for(int i=0;i<s.length();++i) { if(wordDict.find(s.substr(0, i+1)) != wordDict.end()) { canBreak[i] = true; continue; } for(int pre=0;pre<i;++pre) { if(canBreak[pre] && wordDict.find(s.substr(pre+1, i-pre)) != wordDict.end()) { canBreak[i] = true; break; } } } return canBreak[s.length()-1]; } };
https://leetcode.com/problems/word-break-ii/
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
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"].
class Solution { public: void findBreakPoint(vector<bool>& canBreak, string& s, unordered_set<string>& wordDict) { for(int i=0;i<s.length();++i) { if(wordDict.find(s.substr(0, i+1)) != wordDict.end()) { canBreak[i] = true; continue; } for(int pre=0;pre<i;++pre) { if(canBreak[pre] && wordDict.find(s.substr(pre+1, i-pre)) != wordDict.end()) { canBreak[i] = true; break; } } } } void dfs(vector<string>& res, vector<string>& load, vector<bool>& canBreak, string& s, unordered_set<string>& wordDict, int idx) { if(idx == s.length()-1) { string tmp = ""; for(int i=0;i<load.size()-1;++i) tmp += load[i] + " "; if(load.size()>0) tmp+=load[load.size()-1]; res.push_back(tmp); return; } for(int nx=idx+1;nx<s.length();++nx) { if(canBreak[nx] && wordDict.find(s.substr(idx+1, nx-idx)) != wordDict.end()) { load.push_back(s.substr(idx+1, nx-idx)); dfs(res, load, canBreak, s, wordDict, nx); load.pop_back(); } } } vector<string> wordBreak(string s, unordered_set<string>& wordDict) { vector<bool> canBreak(s.length(), false); vector<string> res; res.clear(); vector<string> load; load.clear(); findBreakPoint(canBreak, s, wordDict); if(canBreak[s.length()-1]) dfs(res, load, canBreak, s, wordDict, -1); return res; } };
leetcode@ [139/140] Word Break & Word Break II
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原文地址：http://www.cnblogs.com/fu11211129/p/4999001.html
