[leetcode]Word Break 先上自己的代码 1 class Solution { 2 public: 3 bool wordBreak(string s, unordered_set<string> &dict) { 4 vector<bool> dp(s.size()+1,false);//dp[i]表示从下标0开始的长度为i的子串能否满足word break; 5 dp[0] = true;//相当于分割成了空串和完整的字符串s。不...
Word Break II (LeetCode) Question: https://oj.leetcode.com/problems/word-break-ii/ 解答: 跟Word Break类似,只不过把bool变量变成vector<int>,保存所有能够跟当前点形成一个有效word的点,并且那个点本身的vector<int>不为空(等同于bool flag = true)。所以我们从点(n-1)开始回溯,就能遍历所有点。 比如...
Can you solve this real interview question? Word Break II - Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note
Leetcode: Word Break Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given s ="leetcode", dict =["leet","code"]. Return true because"leetcode"can be segmented as"leet cod...
Leetcode: Word Break 题目: Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given s = “leetcode”, dict = [“leet”, “code”]....
(wordList): if word == endWord: end_index = i break # 如果结束单词不在单词列表中,则无法转换,直接返回 0 if end_index == -1: return 0 # 构建邻接表 adj: Dict[str, List[int]] = defaultdict(list) for i, word in enumerate(wordList): # 枚举 word 替换的字符 for j in range(len(...
Return true because "leetcode" can be segmented as "leet code". ** My code: importjava.util.Set;publicclassSolution{publicbooleanwordBreak(Strings,Set<String>wordDict){boolean[]isBreakUp=newboolean[s.length()+1];isBreakUp[0]=true;for(inti=1;i<s.length()+1;i++){for(intj=0;j<i;j...
其实思路很简单。就是dfs + memory cache reference: https://discuss.leetcode.com/topic/27855/my-concise-java-solution-based-on-memorized-dfs/3 time complexity: O(len(wordDict) ^ len(s / minWordLenInDict)) 然后BFS的做法不知道怎么做。
本周的算法题是 LeetCode 的139.Work Break和140.Work Break II. 我本来只想做一下 140 题,这道题目的要求简单来说就是,输入一个字符串s和一个由不同单词组成的字典wordDict,由你来判断上面输入的s是否可以通过只添加空格的方式(只能把字符串拆分成多个“单词”,但是不能调整字符顺序)拆分成由wordDict中的...
break old_dp = new_dp[:] ans = 0 for i in range(12 * n, 20 * n + 1): ans = (new_dp[i] + ans) % MOD # print(ans) return str(ans) if __name__ == "__main__": # You can add more test cases here print(solution(3) == "19195617" ) ...