[Leetcode] Word Break II

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"].

这题不能直接DFS，否则会超时，联想到上一题，可以跟上题一样先动态规划，判断能否被break，如果s不能被break，那么也没有DFS的必要了，另外在DFS时也可以再利用dp所存的信息从而可以大大得剪掉不必要的操作。

 1 class Solution {

 2 public:

 3     void breakWord(vector<string> &res, string &s, unordered_set<string> &dict, string str, int idx, vector<bool> &dp) {

 4         string substr;

 5         for (int len = 1; idx + len < s.length() + 1; ++len) {

 6             if (dp[idx + len] && dict.count(s.substr(idx,len)) > 0) {

 7                 substr = s.substr(idx, len);

 8                 if (idx + len < s.length()) {

 9                     breakWord(res, s, dict, str + substr + " ", idx + len, dp);

10                 } else {

11                     res.push_back(str + substr);

12                     return;

13                 }

14             }

15         }

16     }

17     

18     vector<string> wordBreak(string s, unordered_set<string> &dict) {

19         vector<bool> dp(s.length() + 1, false);

20         dp[0] = true;

21         for (int i = 0; i < s.length(); ++i) {

22             if (dp[i]) {

23                 for (int len = 1; i + len < s.length() + 1; ++len) {

24                     if (dict.count(s.substr(i, len)) > 0) {

25                         dp[i + len] = true;

26                     }

27                 }

28             }

29         }

30         vector<string> res;

31         if (dp[s.length()]) 

32             breakWord(res, s, dict, "", 0, dp);

33         return res;

34     }

35 };