LeetCode-Word Break II-优快云博客

本文链接：https://blog.youkuaiyun.com/HolyDominat/article/details/38271307

本文探讨了如何将字符串 s 分割成一系列字典中有效的词汇，通过深度优先搜索和动态规划算法实现字符串匹配与句子构建。

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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"].

Solution:

Code:

<span style="font-size:14px;">class Solution {
public:
    void DFS(vector<string> &results, const vector<vector<int> > &dp, int index, string result, const string &s) {
        const int length = dp[index].size();
        if (length == 0) return;
        for (int i = 0; i < length; ++i) {
            const int begin = dp[index][i];
            if (begin == 0)
                results.push_back(s.substr(0, index+1)+result);
            else
                DFS(results, dp, begin-1, " "+s.substr(begin, index-begin+1)+result, s);
        }
    }
    
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        vector<string> results;
        const int length = s.size();
        vector<vector<int> > dp(length, vector<int>());
        for (int i = 0; i < length; ++i) {
            if (dict.find(s.substr(0, i+1)) != dict.end())
                dp[i].push_back(0);
            for (int j = 0; j < i; ++j)
                if (!dp[j].empty() && dict.find(s.substr(j+1, i-j)) != dict.end())
                    dp[i].push_back(j+1);
        }
        DFS(results, dp, length-1, "", s);
        return results;
    }
};</span>