word-break-ii

最新推荐文章于 2019-07-01 11:51:10 发布

njudongchen

最新推荐文章于 2019-07-01 11:51:10 发布

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分类专栏： leetcode 文章标签： string

本文链接：https://blog.youkuaiyun.com/njudongchen/article/details/77689291

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本文介绍了一种基于记忆化搜索的字符串拆分算法，该算法能够有效地将一个字符串通过给定的字典拆分成多个合法词汇组成的句子。文章通过示例说明了如何利用此算法找到所有可能的拆分方式。

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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”].

记忆化搜索，将对应位置之后的所有的string保存下来，供下一次递归使用，减少计算量

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        int max_len = 0;
        for (auto c : dict)
        {
            if (c.size()>max_len)
                max_len = c.size();
        }
        int flag = 1;
        unordered_map<int, vector<string>> map;
        vector<string> res = recurfind(s, dict, 0, max_len, flag, map);
        return res;
    }

    vector<string>  recurfind(string &s, unordered_set<string> &dict, int beg, int max_len, int &flag, unordered_map<int, vector<string>> &map){
        if (flag == 0)
            return vector<string>{};
        if (beg >= s.size() && flag)
            return vector <string> {string()};

        vector<string> vec;
        flag = 0;
        for (int i = 1; i <= max_len&&beg + i <= s.size(); ++i)
        {
            if (dict.count(s.substr(beg, i))){
                flag = 1;
                string str = s.substr(beg, i) + " ";
                vector<string> res;
                if (map.count(beg + i))
                    res = map[beg + i];
                else
                    res = recurfind(s, dict, beg + i, max_len, flag, map);
                if (flag == 0)
                {
                    continue;
                }
                for (auto c : res){
                    if (c.size() == 0)
                        c.insert(0, str.substr(0, str.size() - 1));
                    else
                        c.insert(0, str);
                    vec.push_back(c);
                }
            }
        }
        map[beg] = vec;
        return vec;
    }
};