算法设计与应用基础

xiaoxiao2021-02-27  676

139. Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given s = "leetcode", dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

bool wordBreak(string s, unordered_set<string> &dict) { if(dict.size()==0) return false; vector<bool> dp(s.size()+1,false); dp[0]=true; for(int i=1;i<=s.size();i++) { for(int j=i-1;j>=0;j--) { if(dp[j]) { string word = s.substr(j,i-j); if(dict.find(word)!= dict.end()) { dp[i]=true; break; //next i } } } } return dp[s.size()]; }
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