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: []
Approach #1: Recursive. [C++]
class Solution {
private:
unordered_map<string, vector<string>> m;
vector<string> combine(string word, vector<string> prev) {
for (int i = 0; i < prev.size(); ++i) {
prev[i] += ' ' + word;
}
return prev;
}
public:
vector<string> wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> wordDict_(wordDict.begin(), wordDict.end());
if (m.count(s)) return m[s];
vector<string> result;
if (wordDict_.find(s) != wordDict_.end()) {
result.push_back(s);
}
for (int i = 1; i < s.size(); ++i) {
string word = s.substr(i);
if (wordDict_.find(word) != wordDict_.end()) {
string rem = s.substr(0, i);
vector<string> prev = combine(word, wordBreak(rem, wordDict));
result.insert(result.end(), prev.begin(), prev.end());
}
}
m[s] = result;
return result;
}
};