Java for LeetCode 140 Word Break II】的更多相关文章

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", dict = ["cat", "cats&quo…
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. Note: The same word in the dictionary m…
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, givens = "catsanddog",dict = ["cat", "cats"…
原题地址 动态规划题 令s[i..j]表示下标从i到j的子串,它的所有分割情况用words[i]表示 假设s[0..i]的所有分割情况words[i]已知.则s[0..i+1]的分割情况words[i+1] = words[k] + s[k+1..i+1],其中(有三个条件要满足)(1) 0 <= k <= i,(2) words[k]非空,(3) s[k+1..i+1]在字典中. 根据这个递推公式求解,有两种枚举方式: 1. 对于每个待求解的位置i,从0到i枚举所有的k,然后检验words[…
139. Word Break 字符串能否通过划分成词典中的一个或多个单词. 使用动态规划,dp[i]表示当前以第i个位置(在字符串中实际上是i-1)结尾的字符串能否划分成词典中的单词. j表示的是以当前i的位置往前找j个单词,如果在j个之前能正确分割,那只需判断当前这j单词能不能在词典中找到单词.j的个数不能超过词典最长单词的长度,且同时不能超过i的索引. 初始化时要初始化dp[0]为true,因为如果你找第一个刚好匹配成功的,你的dp[i - j]肯定就是dp[0].因为多申请了一个,所以d…
欢迎fork and star:Nowcoder-Repository-github 140. Word Break II 题目: Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may a…
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", dict = ["cat",…
Word Break II 题解 题目来源:https://leetcode.com/problems/word-break-ii/description/ Description Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid d…
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: Only one letter can be changed at a time    Each intermediate word must exist in the dictionary For example, Given:start = "…
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, givens = "catsanddog",dict = ["cat", &q…