[Leetcode] 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".

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。

class Solution {
public:
    /**
     * @param s: A string s
     * @param dict: A dictionary of words dict
     */
    bool wordBreak(string s, unordered_set<string> &dict) {
        // write your code here
        vector<bool> dp(s.length() + 1, false);
        dp[0] = true;
        int min_len = INT_MAX, max_len = INT_MIN;
        for (auto &ss : dict) {
            min_len = min(min_len, (int)ss.length());
            max_len = max(max_len, (int)ss.length());
        }
        for (int i = 0; i < s.length(); ++i) if(dp[i]) {
            for (int len = min_len; i + len <= s.length() && len <= max_len; ++len) {
                if (dict.find(s.substr(i, len)) != dict.end()) 
                    dp[i + len] = true;
            }
            if (dp[s.length()]) return true;
        }
        return dp[s.length()];
    }
};

再看个非动规的版本：

class Solution {
public:
  bool wordBreakHelper(string s,unordered_set<string> &dict,set<string> &unmatched,int mn,int mx) {  
        if(s.size() < 1) return true;  
        int i = mx < s.length() ? mx : s.length();  
        for(; i >= mn ; i--)  
        {  
            string preffixstr = s.substr(0,i);  
            if(dict.find(preffixstr) != dict.end()){  
                string suffixstr = s.substr(i);  
                if(unmatched.find(suffixstr) != unmatched.end())  
                    continue;  
                else  
                    if(wordBreakHelper(suffixstr, dict, unmatched,mn,mx))  
                        return true;  
                    else  
                        unmatched.insert(suffixstr);  
            }  
        }  
        return false;  
    }  
    bool wordBreak(string s, unordered_set<string> &dict) {  
        // Note: The Solution object is instantiated only once.  
        if(s.length() < 1) return true;  
        if(dict.empty()) return false;  
        unordered_set<string>::iterator it = dict.begin();  
        int maxlen=(*it).length(), minlen=(*it).length();  
        for(it++; it != dict.end(); it++)  
            if((*it).length() > maxlen)  
                maxlen = (*it).length();  
            else if((*it).length() < minlen)  
                minlen = (*it).length();  
        set<string> unmatched;  
        return wordBreakHelper(s,dict,unmatched,minlen,maxlen);  
    }
};