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| 1 | +// Source : https://leetcode.com/problems/word-break-ii/ |
| 2 | +// Author : henrytien |
| 3 | +// Date : 2022年02月03日 |
| 4 | + |
| 5 | +/***************************************************************************************************** |
| 6 | + * |
| 7 | + * Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence |
| 8 | + * where each word is a valid dictionary word. Return all such possible sentences in any order. |
| 9 | + * |
| 10 | + * Note that the same word in the dictionary may be reused multiple times in the segmentation. |
| 11 | + * |
| 12 | + * Example 1: |
| 13 | + * |
| 14 | + * Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"] |
| 15 | + * Output: ["cats and dog","cat sand dog"] |
| 16 | + * |
| 17 | + * Example 2: |
| 18 | + * |
| 19 | + * Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"] |
| 20 | + * Output: ["pine apple pen apple","pineapple pen apple","pine applepen apple"] |
| 21 | + * Explanation: Note that you are allowed to reuse a dictionary word. |
| 22 | + * |
| 23 | + * Example 3: |
| 24 | + * |
| 25 | + * Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"] |
| 26 | + * Output: [] |
| 27 | + * |
| 28 | + * Constraints: |
| 29 | + * |
| 30 | + * 1 <= s.length <= 20 |
| 31 | + * 1 <= wordDict.length <= 1000 |
| 32 | + * 1 <= wordDict[i].length <= 10 |
| 33 | + * s and wordDict[i] consist of only lowercase English letters. |
| 34 | + * All the strings of wordDict are unique. |
| 35 | + ******************************************************************************************************/ |
| 36 | + |
| 37 | +#include "../inc/ac.h" |
| 38 | +class Solution |
| 39 | +{ |
| 40 | +public: |
| 41 | + vector<string> wordBreak(string s, vector<string> &wordDict) |
| 42 | + { |
| 43 | + unordered_map<int, vector<string>> memo{{s.size(), {""}}}; |
| 44 | + |
| 45 | + function<vector<string>(int i)> sentences = [&](int i) |
| 46 | + { |
| 47 | + if (!memo.count(i)) |
| 48 | + { |
| 49 | + for (int j = i + 1; j <= s.size(); j++) |
| 50 | + { |
| 51 | + string token = s.substr(i, j - i); |
| 52 | + if (std::count(wordDict.begin(), wordDict.end(), token)) |
| 53 | + { |
| 54 | + for (auto &&tail : sentences(j)) |
| 55 | + { |
| 56 | + memo[i].push_back(token + (tail == "" ? "" : ' ' + tail)); |
| 57 | + } |
| 58 | + } |
| 59 | + } |
| 60 | + } |
| 61 | + return memo[i]; |
| 62 | + }; |
| 63 | + return sentences(0); |
| 64 | + } |
| 65 | +}; |
| 66 | + |
| 67 | +int main() |
| 68 | +{ |
| 69 | + |
| 70 | + string s = "catsanddog"; |
| 71 | + vector<string> word_dict = {"cat", "cats", "and", "sand", "dog"}; |
| 72 | + // string s = "pineapplepenapple"; |
| 73 | + // vector<string> word_dict = {"apple", "pen", "applepen", "pine", "pineapple"}; |
| 74 | + // string s = "catsandog"; |
| 75 | + // vector<string> word_dict = {"cats", "dog", "sand", "and", "cat"}; |
| 76 | + auto v = Solution().wordBreak(s, word_dict); |
| 77 | + for (auto &&iter : v) |
| 78 | + cout << iter << " "; |
| 79 | + cout << "\n"; |
| 80 | +} |
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