Word Break

Question

http://www.lintcode.com/en/problem/word-break/
Given a string s and a dictionary of words dict, determine if s can be break into a space-separated sequence of one or more dictionary words.

Example

Given s = "lintcode", dict = ["lint", "code"].
Return true because "lintcode" can be break as "lint code".

Answer

Run Time : 582ms 

class Solution {
public:
 /*
 * @param s: A string
 * @param dict: A dictionary of words dict
 * @return: A boolean
 */

 bool wordBreak(string &s, unordered_set<string> &dict) {
  // write your code here
  if (dict.size() <= 0) {
   if (s.length() <= 0)
    return true;
   return false;
  }
  Trie trie;
  for (auto d : dict)
   trie.insert(d);
  return wordBreakHelp(s, trie, 0);
 }
private:
 struct TrieNode {
  TrieNode() {
   for (int i = 0; i < 26; i++) {
    next[i] = 0;    
   }
   s = 0;
  }
  int s;
  TrieNode* next[26];
 };
 class Trie {
 public:
  Trie() {
   root = new TrieNode();
  }
  inline void deleteHelper(TrieNode* root) {
   if (root) {
    for (int i = 0; i < 26; i++) {
     deleteHelper(root->next[i]);
    }
    delete root;
   }
  }
  ~Trie() {
   deleteHelper(root);
   root = 0;
  }
  void insert(string& s) {
   TrieNode* iter = root;
   for (auto c : s) {
    if (iter->next[c - 'a'] == 0)
     iter->next[c - 'a'] = new TrieNode();
    iter = iter->next[c - 'a'];
   }
   iter->s = s.length();
  }
  unordered_set<int> search(string& s, int pos) {
   TrieNode* iter = root;
   unordered_set<int> words;
   for (int i = pos; i < s.length(); i++) {
    char c = s[i];
    if (iter->next[c - 'a'] == 0)
     break;
    if (iter->s > 0) {
     words.insert(iter->s);
    }
    iter = iter->next[c - 'a'];
   }
   if (iter->s > 0) {
    words.insert(iter->s);
   }
   return words;
  }
 private:
  TrieNode* root;
 };
 inline bool wordBreakHelp(string& s, Trie& trie, int pos) {
  unordered_set<int> words = trie.search(s, pos);
  if (s.length() > 0 && words.size() > 0) {
   for (auto word : words) {
    if ((pos + word) == s.length())
     return true;    
    if (wordBreakHelp(s, trie, pos+word))
     return true;
   }
  }
  return false;
 }
};