Leetcode: Word Break II

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"].

原来觉得很复杂的题，现在感觉一般般了，学习使人进步吧~~ DP，思路对了比较简单。

类似于Word Break，用sentences[i]表示i之后所有的解，则最终的解为sentences[0]。

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        int size = s.size();
        if (size == 0 || dict.empty()) {
            return vector<string>();
        }

        vector<vector<string>> sentences(size+1);
        sentences[size].push_back("");
        for (int i = size - 1; i >= 0; --i) {
            for (int j = i; j < size; ++j) {
                string cur = s.substr(i, j - i + 1);
                if (dict.find(cur) != dict.end() && !sentences[j+1].empty()) {
                    for (int k = 0; k < sentences[j+1].size(); ++k) {
                        if (sentences[j+1][k].empty()) {
                            if (j + 1 == size) {
                                sentences[i].push_back(cur);
                            }
                        }
                        else {
                            sentences[i].push_back(cur + " " + sentences[j+1][k]);
                        }
                    }
                }
            }
        }
        
        return sentences[0];
    }
};

================第二次================

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        int size = s.size();
        vector<vector<string>> result(size+1, vector<string>());
        for (int i = size - 1; i >= 0; --i) {
            for (int j = i; j < size; ++j) {
                string word = s.substr(i, j-i+1);
                if (dict.find(word) != dict.end()) {
                    if (j + 1 >= size) {
                        result[i].push_back(word);
                    }
                    else {
                        for (int k = 0; k < result[j+1].size(); ++k) {
                            result[i].push_back(word + " " + result[j+1][k]);
                        }
                    }
                }
            }
        }
        
        return result[0];
    }
};