LeetCode -- Word Break 动态规划,详细理解 Given a stringsand a dictionary of wordsdict, determine ifscan 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...
题目地址:https://leetcode.com/problems/word-break-ii/ 题目解析:看到题目的第一思路是采用递归暴力解法,每找到一个单词将单词添加到返回的结果集中,并将查找的开始位置后移直到字符串的结尾。 题目解答: import java.util.HashSet; import java.util.LinkedList; import java.util.List; import java.util.Set...
【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. For example, given s = “catsanddog...
s ="leetcode", dict =["leet", "code"]."leetcode"can be segmented as"leet code". 思路: res[i] = res[j]&wordDict.contains(s.subString(j,i)); 从0~i的字符串能用单词表示 取决于 是否存在从0~j的字符串能用单词表示,且从j~i是一个单词。j<i 算法: 1. public boolean wordBreak(Stri...
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
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...
提交后,leetcode提示超时,看到relate topic提示是动态规划方法。 于是从这个角度想到了这个解法,dp[k]表示从头开始到第k个字符的字符串能否被dict完美break,dp[0]默认为true,可理解为空字符串能够被break。 求解dp[k]是否为true,只需要看0<=i<k这个范围是否存在某处使得dp[i] == true并且s.substring(i, k...
Leetcode solution 243: Shortest Word Distance 2019-12-26 03:41 − Problem Statement Given a list of words and two words word1 and word2, return the shortest distance between these two word... 包子模拟面试 0 213 [Leetcode] 58. Length of Last Word 2019-12-10 14:07 − Given...
https://leetcode.com/problems/word-ladder/description/ Problem: Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word mus...
leet s[0:0+4] in wordDict s[0+4] = True l e e t c o d e T F F F T F F F F 当搜索到这里时会再次进行重复的搜索。 --- emmm, 写法待改进。 这个写法思路一样,不过效率会低。 beat 3%. 测试地址: https://leetcode.com/problems/word-break/description/ """ class Solution(...