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…

Word Break 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/word-break/description/ Description 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…

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. Note: The same word in the dictionary may be reused multiple t…

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, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

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…

原题地址 与Word Break II(参见这篇文章)相比,只需要判断是否可行,不需要构造解,简单一些. 依然是动态规划. 代码: bool wordBreak(string s, unordered_set<string> &dict) { ; for (auto w : dict) maxLen = maxLen > w.length() ? maxLen : w.length(); vector<, false); res[s.length()] = true; ;…

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…

题目: Given a non-empty string s and a dictionary wordDict containing a list of non-emptywords, determine if s can be segmented into a space-separated sequence of one or more dictionary words. Note: The same word in the dictionary may be reused multipl…

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"…

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, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

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…

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…

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. Note: The same word in the dictionary may be reused multiple t…

原题地址: https://leetcode.com/problems/word-break/description/ 题目: 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…

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, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

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…

思路是这种.我们从第一个字符開始向后依次找,直到找到一个断句的地方,使得当前获得的子串在dict中,若找到最后都没找到.那么就是False了. 在找到第一个后,接下来找下一个断句处,当然是从第一个断句处的下一个字符開始找连续的子串,可是这时与第一个就稍有不同.比方说word='ab', dict={ 'a', ab', ...},在找到a后,接下来处理的是b.我们发现b不在dict中,可是我们发现b能够和a结合,形成ab,而ab在dict中.所以这里的每一个子串就能够有三种选择.要么自己单独作为…

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. Note: The same word in the dictionary may be reused multiple t…

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 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…

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"…

http://oj.leetcode.com/problems/word-break/ 动态规划题目,重点是建立出模型来: fun(start,end) = fun(start,i)*fun(i+1,end); 二维动态数组的申请: int len = s.length(); int **flag = new int *[len]; for(int i = 0;i<len;i++)    flag[i] = new int [len]; #include <iostream> #incl…

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[…

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…

题目地址:请戳我 这一题在leetcode前面一道题word break 的基础上用数组保存前驱路径,然后在前驱路径上用DFS可以构造所有解.但是要注意的是动态规划中要去掉前一道题的一些约束条件(具体可以对比两段代码),如果不去掉则会漏掉一些解(前一道题加约束条件是为了更快的判断是字符串是够能被分词,这里是为了找出所有分词的情况) 代码如下: class Solution { public: vector<string> wordBreak(string s, unordered_set<…

原题链接在这里:https://leetcode.com/problems/word-break-ii/ 题目: 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 = "c…

Problem link: http://oj.leetcode.com/problems/word-break-ii/ This problem is some extension of the word break problem, so the solution is based on the discussion in Word Break. We also use DP to solve the problem. In this solution, A[i] is not a bool…

 1. Word Break 题目链接 题目要求: 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&q…

原题地址:https://oj.leetcode.com/problems/word-break-ii/ 题意: 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 = "c…