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
HashSet<String> set =newHashSet<String>(); LinkedList<String> sb =newLinkedList<String>(); HashMap<Integer,HashSet<Integer>> map=newHashMap<Integer,HashSet<Integer>>();publicList<String> wordBreak(String s , Set<String>dict){ detected=newboolean[s.length()+1];boolean[] result =newbool...
1. Using the dynamic programming method to find all possible position to cut the string(The same as Word Break) This step will return a boolean matrix, if(i,j) entry is true then the substring s.substring(i,j) can be splitted. 2. Using the boolean matrix obtained in step1, we do ...
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...
原题链接: http://oj.leetcode.com/problems/word-break-ii/ 这道题目要求跟Word Break比较类似,不过返回的结果不仅要知道能不能bre...[LeetCode] 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 wor...
if (i == 0 || pos[i] != null) {// 0~i-1之间字符串可break for (int j = i + 1; j <= s.length(); j++) {// 剩余字符串i~j String sub = s.substring(i, j); if (wordDict.contains(sub)) {// 找到下一个单词
Word Break II -- LeetCode 原题链接:http://oj.leetcode.com/problems/word-break-ii/ 这道题目要求跟Word Break比较类似,不过返回的结果不仅要知道能不能break,如果可以还要返回所有合法结果。一般来说这种要求会让动态规划的效果减弱很多,因为我们要在过程中记录下所有的合法结果,中间的操作会使得算法的复杂度...
(self, s, dict, stringlist): if self.check(s, dict): if len(s) == 0: Solution.res.append(stringlist[1:]) for i in range(1, len(s)+1): if s[:i] in dict: self.dfs(s[i:], dict, stringlist+' '+s[:i]) def wordBreak(self, s, dict): Solution.res = [] self.dfs...
2. Word Break II 题目链接 题目要求: Given a stringsand a dictionary of wordsdict, add spaces insto construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, given s="catsanddog", ...
publicbooleanwordBreak(Strings,Set<String>dict){intlen=s.length();boolean[]dp=newboolean[len+1];dp[0]=true;// 当前字符串:0~ifor(inti=1;i<=len;i++){// 将当前的字符串再进行拆分,查看它是否包含在 dict 中for(intj=0;j<i;j++){// dp[j] 是前面已经计算出来的结果if(dp[j]&&dict....