【[LeetCode] 139 Word Break（BFS统计层数的方法）】的更多相关文章

[LeetCode] 139 Word Break（BFS统计层数的方法）

原题地址: https://leetcode.com/problems/word-break/description/ 题目: Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words…

[LeetCode] 139. Word Break 单词拆分

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. Note: The same word in the dictionary may be reused multiple t…

leetcode 139. Word Break 、140. Word Break II

139. Word Break 字符串能否通过划分成词典中的一个或多个单词. 使用动态规划,dp[i]表示当前以第i个位置(在字符串中实际上是i-1)结尾的字符串能否划分成词典中的单词. j表示的是以当前i的位置往前找j个单词,如果在j个之前能正确分割,那只需判断当前这j单词能不能在词典中找到单词.j的个数不能超过词典最长单词的长度,且同时不能超过i的索引. 初始化时要初始化dp[0]为true,因为如果你找第一个刚好匹配成功的,你的dp[i - j]肯定就是dp[0].因为多申请了一个,所以d…

Leetcode#139 Word Break

原题地址 与Word Break II(参见这篇文章)相比,只需要判断是否可行,不需要构造解,简单一些. 依然是动态规划. 代码: bool wordBreak(string s, unordered_set<string> &dict) { ; for (auto w : dict) maxLen = maxLen > w.length() ? maxLen : w.length(); vector<, false); res[s.length()] = true; ;…

LeetCode 139. Word Break单词拆分 (C++)

题目: Given a non-empty string s and a dictionary wordDict containing a list of non-emptywords, determine if s can be segmented into a space-separated sequence of one or more dictionary words. Note: The same word in the dictionary may be reused multipl…

leetcode 139. Word Break ----- java

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, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

LeetCode #139. Word Break C#

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]139. Word Break单词能否拆分

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. Note: The same word in the dictionary may be reused multiple t…

[LeetCode] 139. Word Break 拆分词句

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. Note: The same word in the dictionary may be reused multiple t…

Java for LeetCode 139 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…