A solution is["cats and dog", "cat sand dog"]. 题解: 这道题不仅仅是看是不是wordbreak,还需要在此基础上把所有word break的结果保存。 为了把所有可能性都保存,那么就使用DFS方法来解决。DFS主要就是跳的层次不容易看出,我下面就以字符串leetcode字典le l et eet code作为例子画了一张图,大概讲解了...
最后就是在LeetCode中,有一个复杂的用例,直接用上面的也会TLE,需要先进行Word Break中的判断,是否可以Break,如果可以则用上面的获取即可,AC的代码如下: publicclassSolution {publicList<String> wordBreak(String s, Set<String>wordDict) {boolean[] isBreak =newboolean[s.length() + 1]; isBreak[0] =tr...
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 code". 动态规划...
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 140. Word Break II(java) 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 such possible ......
LeetCode Top 100 Liked Questions 139. Word Break (Java版; Medium) 题目描述 AI检测代码解析 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 140 word break Solution.php <?php class Solution { /** * @param String $s * @param String[] $wordDict * @return String[] */ public function wordBreak($s, $wordDict) { $n = strlen($s); $dp = new SplFixedArray($n + 1);...
break; } } } return dp[s.length()]; } } Word Break II 链接:https://leetcode.com/problems... 和上一题不同的是,这道题要返回所有可能的组合。所以现在dp[i]里面应该存可以使长度为i所有可能的String里的最后一个word。dp有两种写法,一个就是直接写成数组:List[]的形式,不能形成的dp[i] = ...
怎样避免Leetcode 318题中的常见错误? 文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution 解析: Version 1每次迭代都需要进行两次set转换,Version 2预先进行了set转换,效率有提升。由于Version 2中两个字符串的与运算非常耗时,因此Version 3先将字符串转换为代表字符串的数字,然后进行...
LeetCode :Word Ladder II My Solution Given two words (startandend), and a dictionary, find all shortest transformation sequence(s) fromstarttoend, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary...