LeetCode -- Word Break 动态规划,详细理解 Given a stringsand a dictionary of wordsdict, determine ifscan be segmented into a space-separated sequence of one or more dictionary words. For example, given s="leetcode", dic
http://oj.leetcode.com/problems/word-break-ii/ This problem is some extension of the word break problem, so the solution is based on the discussion inWord Break. We also use DP to solve the problem. In this solution, A[i] is not a boolean any more, but a list of all possible va...
【Leetcode】Word Break II 题目链接:https://leetcode.com/problems/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...
140. Word Break II Hard Topics Companies 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 that the same word in the dictionary may be reused...
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”]....
参考了Discuss区一个DFS+DP的解法(https://leetcode.com/problems/word-break-ii/discuss/44262/My-C%2B%2B-DP%2BDFS-solution-(4ms)),学到了一种新的剪枝策略: classSolution{private://DFS path build functionvoidbuildPath(boolisBreakable[],string&s,intpos,vector<string>&res,string curP,unordered_...
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...
https://leetcode.com/problems/word-ladder/description/ Problem: Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word mus...
[leetcode] 140. Word Break II Description Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all su...[leetcode]140. Word Break II(Java) ......
bool wordBreak(string s, unordered_set<string>& wordDict) { const int n = s.length(); size_t max_len = 0; for (const auto& str: wordDict) { max_len = max(max_len, str.length()); } vector<bool> canBreak(n + 1, false); canBreak[0] = true; for (int i = 1; i <=...