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
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...
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...
classSolution {public: vector<string> wordBreak(strings, unordered_set<string> &dict) { vector<vector<int>> flag(s.size() +1, vector<int>()); flag[0].push_back(0);for(inti =1; i <= s.size(); i++) //当前判断字符串的结束位置的下一个位置 如i=1 表示结束位置是 s[0] 从前向...
LeetCode 140. Word Break II(dfs) 题目来源:https://leetcode.com/problems/word-break-ii/ 问题描述 140. Word Break II Hard Given a non-empty string s and a dictionary wordDict containing a list of non-emp...[leetcode] 140. Word Break II Description Given a non-empty string s and...
【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....
vector<string> wordBreak(string s, vector<string>& wordDict) { unordered_set<string> dict(wordDict.begin(),wordDict.end()); vector<bool> possible(s.size()+1,true); vector<string> ans; string item; dfs(s,0,dict,possible,ans,item); ...
140. Word Break II 题目描述(困难难度) 139 题的升级版。给一个字符串,和一些单词,找出由这些单词组成该字符串的所有可能,每个单词可以用多次,也可以不用。 完全按照139 题的思路做了,大家可以先过去看一下。 解法一 动态规划 先考虑139 题最后一个解法,动态规划,看起来也比较简单粗暴。
126*. Word Ladder II windliang 互联网行业 开发工程师 来自专栏 · LeetCode刷题 题目描述(困难难度) 给定两个单词,一个作为开始,一个作为结束,还有一个单词列表。然后依次选择单词,只有当前单词到下一个单词只有一个字母不同才能被选择,然后新的单词再作为当前单词,直到选到结束的单词。输出这个的...
代码: importjava.util.ArrayList;importjava.util.Set;publicclassSolution{privateArrayList<String>resultList;privateArrayList<String>tmpList;publicArrayList<String>wordBreak(Strings,Set<String>dict){resultList=newArrayList();tmpList=newArrayList();divide(s,dict);returnresultList;}privatevoiddivide(Strings,Set...