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…

class Solution { public: bool wordBreak(string s, vector<string> wordDict) { vector<, false); wordB[] = true; ; i < s.length() + ; i++) { ; j >= ; j--) { if (wordB[j] && find(wordDict.begin(), wordDict.end(), s.substr(j, i - j)) !=…

题目: 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 bec…

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词. 说明: 拆分时可以重复使用字典中的单词.你可以假设字典中没有重复的单词.示例 1: 输入: s = "leetcode", wordDict = ["leet", "code"]输出: true解释: 返回 true 因为 "leetcode" 可以被拆分成 "leet code&qu…

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词. 说明: 拆分时可以重复使用字典中的单词. 你可以假设字典中没有重复的单词. 示例 1: 输入: s = "leetcode", wordDict = ["leet", "code"] 输出: true 解释: 返回 true 因为 "leetcode" 可以被拆分成 "leet cod…