Leetcode -- word break

A person who has never tasted all his wines will not experience the smell of water ~

Given a stringSAnd a dictionary of wordsDict, Determine ifSCan 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"leet code".

 

[Solutions]

DP: ANS [I] indicates the subscript 0 ~ of S ~ Can I be represented by dict?

Ans [I] = the existence of K satisfies ans [k] & dict. Contains (S. substr (k + 1, I-k) = true.

 1 bool Solution::wordBreak(std::string s, std::tr1::unordered_set<std::string> &dict) 2 { 3     if (s.length() == 0 || dict.size()==0) 4     { 5         return false; 6     } 7     int length = s.length(); 8     bool ans[length]; 9     for (int i = 0; i < length; i++)10         ans[i] = false;11     for (int i = 0; i < length; i++)12     {13         if (dict.find(s.substr(0, i+1)) != dict.end())14             ans[i] = true;15         for (int j = 0; j < i; j++)16         {17             if (ans[j]&&dict.find(s.substr(j+1, i-j))!=dict.end())18             {19                 ans[i] = true;20                 break;21             }22         }23     }24     return ans[length-1];25 }