[LeetCode 解题报告]140. Word Break II

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

考察：dfs； 

class Solution {
public:
    vector wordBreak(string s, vector& wordDict) {
        
        vector res;
        string out;
        vector possible(s.size()+1, true);
        
        unordered_set ss(wordDict.begin(), wordDict.end());
        wordBreakDFS(s, ss, 0, possible, out, res);
        return res;
    }
    
    void wordBreakDFS(string &s, unordered_set& ss, int start, vector& possible,
                     string &out, vector& res) {
        if (start == s.size()) {
            res.push_back(out.substr(0, out.size()-1));
            return ;
        }
        
        for (int i = start; i < s.size(); i ++) {
            string word = s.substr(start, i - start + 1);
            if (ss.find(word) != ss.end() && possible[i+1]) {
                out.append(word).append(" ");
                int oldSize =  res.size();
                wordBreakDFS(s, ss, i + 1, possible, out, res);
                if (res.size() == oldSize)
                    possible[i+1] = false;
                out.resize(out.size() - word.size()-1);
            }
        }
    }
};