This problem can also be solved using DFS or BFS. DFS: 1classSolution {2public:3boolwordBreak(strings, unordered_set<string>&wordDict) {4for(stringword : wordDict) {5minlen = min(minlen, (int)word.length());6maxlen = max(maxlen, (int)word.length());7}8unordered_map<string,bool>...
1public class Solution {2public boolean wordBreak(String s, Set<String>dict){3boolean[] t =new boolean[s.length()+1];4t[0]=true;//set first to be true, why?5//Because we need initial state67for(inti=0; i<s.length(); i++){8//should continue from match position9if(!t[i])...
break; } } } return 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. C#参考代码: AI检测 public class Solution { public bool WordBreak(string s, ISet<string> dict) { bool[] matches = new bool[s.Length + 1]; matches[0] = true; ...
得到一个pos数组,每一个元素保存的是从上一个word下标到当前下标之间的字符串能组成的word。然后根据上述结果dfs。 算法2: List<String> res = new ArrayList<String>(); public List<String> wordBreak(String s, Set<String> wordDict) { ArrayList<String>[] pos = new ArrayList[s.length()+1]; for ...
public boolean wordBreak(String s, Set<String> wordDict) { boolean[] dp = new boolean[s.length()+1]; Arrays.fill(dp,false); dp[s.length()]=true; // 外层循环递增长度 for(int i = s.length()-1; i >=0 ; i--){ // 内层循环寻找分割点 ...
基于图的DFS: 和BFS一样一般需要一个set来记录访问过的节点,避免重复访问造成死循环; Word XXX 系列面试中非常常见,例如word break,word ladder,word pattern,word search。 Leetcode 341 Flatten Nested List Iterator (339 364) Leetcode 394 Decode String Leetcode 51 N-Queens (I II基本相同) Leetcode 291...
For the purpose of this problem, we define empty string as valid palindrome. 【解答】注意大小写,注意数字和字母都要算有效字符。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 public class Solution { public boolean isPalindrome(String s) { int left = 0; int right = s.length()-1; whil...
0212 Word Search II Go 37.0% Hard 0213 House Robber II Go 40.6% Medium 0214 Shortest Palindrome 32.2% Hard 0215 Kth Largest Element in an Array Go 65.7% Medium 0216 Combination Sum III Go 67.0% Medium 0217 Contains Duplicate Go 61.2% Easy 0218 The Skyline Problem Go 41.5% Hard...
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publicclassSolution{publicbooleanwordBreak(Strings,Set<String>wordDict){returnthis.isBreakUp(s,wordDict,0,s.length());}privatebooleanisBreakUp(Strings,Set<String>wordDict,inthead,intlength){if(head==length)returntrue;for(inti=head;i<s.length();i++){if(wordDict.contains(s.substring(head,i+...