Word Break II

最新推荐文章于 2021-01-02 23:29:20 发布

贪睡的萝卜

最新推荐文章于 2021-01-02 23:29:20 发布

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分类专栏： leetcode 算法编程文章标签：算法 math leetcode

本文链接：https://blog.youkuaiyun.com/u012365383/article/details/48215391

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本文介绍如何使用动态规划算法解决字符串分词问题，即根据给定的字典将输入字符串拆分为合法的单词组成句子。通过实例演示，详细解释了算法实现过程和关键步骤。

Question

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"].

MySolution

class Solution {
    map<string, vector<string>> map_words; // F(i, N)
public:
    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
        /*动态规划解题
        1. 边界条件：s在dic中，返回s
        2. F(0, N) = ||_{i = 1, N}{F(0, i) + F(i, N)}
        */
        
        auto tmp = map_words.find(s); // 加速
        if(tmp != map_words.end())
        {
            return tmp->second;
        }
        
        vector<string> rst;
        if(wordDict.find(s) != wordDict.end())  // 整体一个词
        {
            rst.push_back(s);
        }
        for(int i = 1; i < s.size(); ++i)
        {
            string tmpSub = s.substr(0, i);
            if(wordDict.find(tmpSub) != wordDict.end())    // 开头截取一个词
            {
                auto tmpRst = wordBreak(s.substr(i), wordDict); // 剩余部分的词
                for(auto &tmpR: tmpRst)
                {
                    rst.push_back(tmpSub + " " + tmpR);
                }
            }
        }
        map_words[s] = rst;
        return rst;
    }
};