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".

动规实在是太强大了！注意在枚举子串长度时，只要枚举从dict字典中最短单词到最长单词的长度就可以了。

1 class Solution { 2 public: 3     /** 4      * @param s: A string s 5      * @param dict: A dictionary of words dict 6      */ 7     bool wordBreak(string s, unordered_set
     
       &dict) { 8         // write your code here 9         vector
      
        dp(s.length() + 1, false);10         dp[0] = true;11         int min_len = INT_MAX, max_len = INT_MIN;12         for (auto &ss : dict) {13             min_len = min(min_len, (int)ss.length());14             max_len = max(max_len, (int)ss.length());15         }16         for (int i = 0; i < s.length(); ++i) if(dp[i]) {17             for (int len = min_len; i + len <= s.length() && len <= max_len; ++len) {18                 if (dict.find(s.substr(i, len)) != dict.end()) 19                     dp[i + len] = true;20             }21             if (dp[s.length()]) return true;22         }23         return dp[s.length()];24     }25 };
      
     

再看个非动规的版本：

1 class Solution { 2 public: 3   bool wordBreakHelper(string s,unordered_set
     
       &dict,set
      
        &unmatched,int mn,int mx) {   4         if(s.size() < 1) return true;   5         int i = mx < s.length() ? mx : s.length();   6         for(; i >= mn ; i--)   7         {   8             string preffixstr = s.substr(0,i);   9             if(dict.find(preffixstr) != dict.end()){  10                 string suffixstr = s.substr(i);  11                 if(unmatched.find(suffixstr) != unmatched.end())  12                     continue;  13                 else  14                     if(wordBreakHelper(suffixstr, dict, unmatched,mn,mx))  15                         return true;  16                     else  17                         unmatched.insert(suffixstr);  18             }  19         }  20         return false;  21     }  22     bool wordBreak(string s, unordered_set
       
         &dict) {  23         // Note: The Solution object is instantiated only once.  24         if(s.length() < 1) return true;  25         if(dict.empty()) return false;  26         unordered_set
        
         ::iterator it = dict.begin();  27         int maxlen=(*it).length(), minlen=(*it).length();  28         for(it++; it != dict.end(); it++)  29             if((*it).length() > maxlen)  30                 maxlen = (*it).length();  31             else if((*it).length() < minlen)  32                 minlen = (*it).length();  33         set
         
           unmatched;  34         return wordBreakHelper(s,dict,unmatched,minlen,maxlen);  35     }36 };