[LeetCode July Challange]Day30-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:
[]

---

solution

time complexity : O(2^n)
space complexity : O(2^n)

class Solution {
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> dict(wordDict.begin(), wordDict.end());
        return wordBreak(s, dict);        
    }
private:
    unordered_map<string, vector<string>> mem_;
    vector<string> append(vector<string> prefixs, string word) {
        for (int i=0; i<prefixs.size(); ++i) {
            prefixs[i] += " " + word;
        }
        return prefixs;
    }
    
    vector<string> wordBreak(string s, unordered_set<string> dict) {
        if (mem_.count(s)) return mem_[s];
        
        vector<string> ans;
        
        if (dict.count(s)) ans.push_back(s);
        
        for (int i=0; i<s.length(); ++i) {
            string right = s.substr(i);
            if (!dict.count(right)) continue;
            string left = s.substr(0, i);
            vector<string> tmpAns = append(wordBreak(left, dict), right);
            ans.insert(ans.end(), tmpAns.begin(), tmpAns.end());
        }
        
        return mem_[s] = ans;
    }
};

利用計劃遞歸的方式完成， 遞歸的終止條件：若此字串s之前已有相對應的答案，則回傳答案。
將此字串s，從左至右，切割成2字串，若右邊字串出現在題目給定的wordDict中的話，左邊的字串遞歸求解，與右字串接在一起。
例：”catsanddog”, dict : {“cat”, “cats”, “sand”, “and”, “dog”}
wordBreak(“catsand”)+” dog”
{wordBreak(“cat”)+” sand”, wordBreak(“cats”)+” and”}+” dog”
{“cat sand”, “cats and”}+” dog”
{“cat sand dog”, “cats and dog”}