[Leetcode] Word Break 1 & 2

Word Break

 

AC Rate: 2/13

My Submissions

 

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".

建字典树搞。

 

class Solution {
public:

    class Node {
    public:
        Node* next[26];
        bool end;
        Node(): end(false) { for (int i = 0; i < 26; i++) next[i] = NULL;}
        void insert(string a) {
            Node * cur = this;
            for (int i = 0; i < a.size(); i++) {
                if (cur->next[a[i]-'a'] == NULL) {
                    cur->next[a[i]-'a'] = new Node();
                }
                cur = cur->next[a[i]-'a'];
            }
            cur->end = true;
        }
        ~Node () {
            for (int i = 0;i < 26; i++) delete next[i];
        }
    };
    
    bool wordBreak(string s, unordered_set<string> &dict) {
        Node root;
        for (auto it = dict.begin(); it != dict.end(); ++it) {
            root.insert(*it);
        }
        
        vector<bool> v(s.size(), false);
        findMatch(s, &root, 0, v);
        for (int i = 0; i < s.size(); i++) 
            if (v[i]) findMatch(s, &root, i+1, v);
        return v[s.size() - 1];
    }
    
    void findMatch(const string& s, Node* cur, int start, vector<bool> &v) {
        int i = start, n = s.size();
        while (i < n) {
            if (cur->next[s[i] - 'a'] != NULL) {
                if (cur->next[s[i] - 'a']->end) v[i] = true;
                cur = cur->next[s[i] - 'a'];
            }
            else break;
            i++;
        }
        
    }
};

 

粗暴一点也是ok的。

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        int n = dict.size();
        int maxlen = 0;
        for (auto it = dict.begin(); it != dict.end(); ++it) 
            if (it->size() > maxlen)   maxlen = it->size();
        int sn = s.size();
        vector<bool> v(sn, false);
        for (int i = 0; i < sn; i++) {
            if (i == 0 || (i > 0 && v[i-1])) {
                for (int j = 1; j <= maxlen && i + j - 1 < sn; j++) {
                    if (dict.count(s.substr(i,j)) > 0) v[i+j-1] = true;
                }
            }
        }
        return v[sn-1];
    }
};

 

 

Word Break II

 

AC Rate: 9/89

My Submissions

 

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:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        vector<string> res;
        int dn = dict.size();
        int sn = s.size();
        int maxlen = 0;
        for (auto it = dict.begin(); it != dict.end(); ++it) {
            if (it->size() > maxlen) maxlen = it->size();
        }
        vector<vector<int> > next(sn);
        vector<bool> v(sn, false);
        for (int i = 0; i < sn; i++) {
            if (i == 0 || (i > 0 && v[i-1])) {
                for (int j = 1; j <= maxlen && i+j-1 < sn; j++) {
                    if (dict.count(s.substr(i, j)) > 0) {
                        v[i+j-1] = true;
                        next[i].push_back(j);
                    }
                }
            }
        }
        if (!v[sn-1]) return res;
        vector<int> x;
        x.push_back(0);
        gen(s, res, next, x);
        
        return res;
    }
    
    void gen(const string& s, vector<string>& res, vector<vector<int> >& next, vector<int> &v) {
        int cur = v.back();
        if (cur == s.size()) {
            string t = s.substr(v[0], v[1] - v[0]);
            for (int i = 1; i < v.size() - 1; i++)
                t += " " + s.substr(v[i], v[i+1] - v[i]);
            res.push_back(t);
            return;
        }
        
        for (int i = 0; i < next[cur].size(); i++) {
            v.push_back(cur+next[cur][i]);
            gen(s, res, next, v);
            v.pop_back();
        }
    }
};