LeetCode -- Word Break 动态规划，详细理解

【LeetCode -- Word Break 动态规划，详细理解】的更多相关文章

LeetCode -- Word Break 动态规划，详细理解

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 ||& Word Break && Word Break II（转）——动态规划

一. 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&…

LeetCode:Word Break II（DP）

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<…

[leetcode]Word Break II @ Python

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…

[LeetCode] Word Break II 拆分词句之二

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…

LeetCode Word Break II

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…