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
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...
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: For example, given s="catsanddog...
}publicstaticvoidmain(String[] args) {//TODO Auto-generated method stubWordBreak wb =newWordBreak();//String s = "bccdbacdbdacddabbaaaadababadad";String s = "catsandcatsand";//String []d = new String[]{"cbc","bcda","adb","ddca","bad","bbb","dad","dac","ba","aa","bd",...
【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....
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...
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); ...
publicList<String>wordBreak(Strings,List<String>wordDict){//提前进行一次判断HashSet<Character>set2=newHashSet<>();for(inti=0;i<wordDict.size();i++){Stringt=wordDict.get(i);for(intj=0;j<t.length();j++){set2.add(t.charAt(j));}}for(inti=0;i<s.length();i++){if(!set2.con...
Word Ladder II - LeetCodeleetcode.com/problems/word-ladder-ii/description/ 根据题意,要找到 beginWord 到 endWord 的所有变换路径,双向搜索的思路是:依次填充以 beginWord 和 endWord 为终点的两个有向图,当两个有向图出现共同单词时,以共同单词为连接点,连接两个有向图中指向终点的路径,就得到 beginWord...
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...