LeetCode 0140 - Word Break II

Reck Zhang

发布于 2021-08-11 14:47:26

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发布于 2021-08-11 14:47:26

文章被收录于专栏：Reck Zhang

Word Break II

Desicription

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

class Solution {
private:
    unordered_map<string, vector<string>> mp; 
    unordered_set<string> dict;
    vector<string> combine(string s, vector<string> vec) {
        for(int i = 0; i < vec.size(); i++)
            vec[i] += " " + s;
        return vec;
    }
    vector<string> dfs(string s) {
        if(mp.count(s))
            return mp[s];
        vector<string> res;
        if(dict.count(s))
            res.push_back(s);
        for(int i = 1; i < s.size(); i++) {
            string word = s.substr(i);
            if(dict.count(word)) {
                string rem = s.substr(0, i);
                vector<string> prev = combine(word, dfs(rem));
                res.insert(res.end(), prev.begin(), prev.end());
            }
        }
        mp[s] = res;
        return res;
    }
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        dict = unordered_set<string>(wordDict.begin(), wordDict.end());
        return dfs(s);
    }
};

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原始发表：2018-05-29，如有侵权请联系 cloudcommunity@tencent.com 删除

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