扫清一切障碍之后,BFS(见解法二代码)和DFS(见解法三代码)就都能实现了。 解法一(Java) classSolution {publicbooleanwordBreak(String s, List<String>wordDict) {boolean[] dp =newboolean[s.length() + 1]; dp[0] =true;for(inti = 1; i <= s.length(); i++) {for(intj = i - 1; j >=...
dict=["leet", "code"]. Return true because"leetcode"can be segmented as"leet code". 题解: 这道题的题解转载自Code ganker,他写的很好。地址:http://blog.csdn.net/linhuanmars/article/details/22358863 “原题链接:http://oj.leetcode.com/problems/word-break/ 这道题仍然是动态规划的题目,我们...
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...
代码语言:java 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(...
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笔记:Word Break 一. 题目描写叙述 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"]....
然而这个优化,对于 leetcode 的tests 并没有什么影响。 让我们继续考虑优化方案,回到之前的图。 假如我们在考虑上图中黄色节点的相邻节点,发现第三层的 abc 在第二层已经考虑过了。所以第三层的 abc 其实不用再考虑了,第三层的 abc 后边的结构一定和第二层后边的结构一样,因为我们要找最短的路径,所以如果产生...
Return true because "leetcode" can be segmented as "leet code". ** My code: importjava.util.Set;publicclassSolution{publicbooleanwordBreak(Strings,Set<String>wordDict){boolean[]isBreakUp=newboolean[s.length()+1];isBreakUp[0]=true;for(inti=1;i<s.length()+1;i++){for(intj=0;j<i;j...
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...
Breadcrumbs leetcode-solutions /cpp / 0139-word-break.cpp Latest commit Cannot retrieve latest commit at this time. HistoryHistory File metadata and controls Code Blame 37 lines (31 loc) · 924 Bytes Raw /* Given a string & dictionary, return true if: Can segment string into 1 or more...