LeetCode 140. Word Break II

最新推荐文章于 2025-04-13 18:39:57 发布

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本文介绍了一种使用回溯法解决字符串拆分问题的算法，该问题要求将输入字符串通过空格拆分成由字典中词汇组成的句子，并返回所有可能的组合结果。文中还探讨了如何改进现有解决方案以降低其复杂度。

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Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

// I currently only do it using backtracking. But this has high complexity..... need to think about how to do it using DP.

void wordBreak(string s, int pos, vector< vector<string > >& res, vector<string>& path, unordered_set<string>& wordDict) {
    if(pos == s.size()) {
        res.push_back(path);
        return;
    }
    for(int i = pos; i < s.size(); ++i) {
        string tmp = s.substr(pos, i - pos + 1);
        if(wordDict.find(tmp) != wordDict.end()) {
            path.push_back(tmp);
            wordBreak(s, i + 1, res, path, wordDict);
            path.pop_back();
        }
    }
}


vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
    vector< vector<string> > res;
    vector<string> path;
    wordBreak(s, 0, res, path, wordDict);
    vector<string> result;
    for(int i = 0; i < res.size(); ++i) {
        string tmp = "";
        for(int j = 0; j < res[i].size(); ++j) {
            tmp = tmp + " " + res[i][j];
        }
        result.push_back(tmp);
    }
    return result;
}


// test
int main(void) {
    string test = "aaaaa";
    unordered_set<string> wordDict{"a", "aaa", "aa", "aaaaa", "aaaa"};

    vector<string> res = wordBreak(test, wordDict);
    for(auto iter : res) {
        cout << iter << endl;
    }
}