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: (Backtracking)
class Solution
{
public:
vector<string> res;
void wordUntil(string s, int n, unordered_set<string> &dict, string result)
{
if (s == "")
{
result.pop_back();
res.push_back(result);
return;
}
for (int i = 1; i <= n; i++)
{
string k = s.substr(0, i);
if (dict.find(k) != dict.end())
{
wordUntil(s.substr(i, n - i), n - i, dict, result + k + " ");
}
}
return;
}
vector<string> wordBreak(string s, vector<string> &wordDict)
{
unordered_set<string> dict;
for (int i = 0; i < wordDict.size(); i++)
{
dict.insert(wordDict[i]);
}
int n = s.length();
string a = "";
wordUntil(s, n, dict, a);
return res;
}
};