https://leetcode.com/problems/word-break/ 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"…

一开始的错误答案与错误思路,幻想直接遍历得出答案: class Solution { public: bool wordBreak(string s, vector<string>& wordDict) { for(int i;i<s.size();i++){ ; for(int j;j<wordDict.size();j++){ if(s.substr(i,wordDict[j].size())==wordDict[j]){ step=wordDict[j].size()…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归求解 日期 题目地址:https://leetcode.com/problems/word-break-ii/ 题目描述 Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to con…

https://leetcode.com/problems/add-and-search-word-data-structure-design/ 本题是在Trie树进行dfs+backtracking操作. Trie树模板代码见:http://www.cnblogs.com/fu11211129/p/4952255.html 题目介绍: Design a data structure that supports the following two operations: void addWord…

Design a data structure that supports the following two operations: void addWord(word) bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter…

problem: Given an input string, reverse the string word by word. For example: Given s = "the sky is blue", return "blue is sky the". 问题分析:如何准确的找到每一个需要清除的空格的位置pos,以及每个word对应的pos范围? 解决方法:需要反转一个字符串,使用int string.find_last_not_of(char c, in…

Design a data structure that supports the following two operations: void addWord(word)bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.…

Design a data structure that supports the following two operations: void addWord(word) bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 正则+统计 日期 题目地址:https://leetcode.com/problems/most-common-word/description/ 题目描述 Given a paragraph and a list of banned words, return the most frequent word that is not in the li…

Design a data structure that supports the following two operations: void addWord(word) bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter…

Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be us…

题目:Given an input string, reverse the string word by word. For example,Given s = "the sky is blue",return "blue is sky the". 要求: 1)首尾有空格的时候,反转后的string要将空格去掉 2)当string有多个连续空格的时候,只保留一个空格. 代码分析: 对多余空格剔除: 思路分析: 1)从原始s 的最末尾开始扫描,如果遇到空格,用whil…

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.The input string does not contain leading or trailing spaces and the words are always separated by a single space.For example,Given s = "t…

Microsoft.Office.Interop.Word 创建word 转载:http://www.cnblogs.com/chenbg2001/archive/2010/03/14/1685746.html 功能总结或者完善. 一.添加页眉 using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Linq; using System…

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters. The input string does not contain leading or trailing spaces and the words are always separated by a single space. For example,Given s =…

题目: Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters. The input string does not contain leading or trailing spaces and the words are always separated by a single space. For example,Given…

LeetCode 81 Search in Rotated Sorted Array II [binary search] <c++> 给出排序好的一维有重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数是否存在,时间复杂度限制\(O(log_2n)\) C++ 因为有重复元素存在,nums[l] <= nums[mid]不能说明[l,mid]区间内一定是单调的,比如数组[1,2,3,1,1,1,1],但是严格小于和严格大于的情况还是可以…

LeetCode 80 Remove Duplicates from Sorted Array II [Array/auto] <c++> 给出排序好的一维数组,如果一个元素重复出现的次数大于两次,删除多余的复制,返回删除后数组长度,要求不另开内存空间. C++ 献上自己丑陋无比的代码.相当于自己实现一个带计数器的unique函数 class Solution { public: int removeDuplicates(std::vector<int>& nums) {…

之前用Aspose.Word进行Word转PDF发现'\'这个字符会被转换成'¥'这样的错误,没办法只能换个方法了.下面是Microsoft.Office.Interop.Word转PDF的方法: public bool WordToPDF(string sourcePath, string targetPath) { bool result = false; Microsoft.Office.Interop.Word.Application application = new Microsof…

Leetcode之二分法专题-275. H指数 II(H-Index II) 给定一位研究者论文被引用次数的数组(被引用次数是非负整数),数组已经按照升序排列.编写一个方法,计算出研究者的 h 指数. h 指数的定义: “h 代表“高引用次数”(high citations),一名科研人员的 h 指数是指他(她)的 (N 篇论文中)至多有 h 篇论文分别被引用了至少 h 次.(其余的 N - h 篇论文每篇被引用次数不多于 h 次.)" 示例: 输入: citations = [0,1,3,5,…

Leetcode之回溯法专题-90. 子集 II(Subsets II) 给定一个可能包含重复元素的整数数组 nums,返回该数组所有可能的子集(幂集). 说明:解集不能包含重复的子集. 示例: 输入: [1,2,2] 输出: [ [2], [1], [1,2,2], [2,2], [1,2], [] ] 分析:是78题的升级版,新增了一些限制条件,nums数组是会重复的,求子集. class Solution { List<List<Integer>> ans = new Arr…

Leetcode之回溯法专题-47. 全排列 II(Permutations II) 给定一个可包含重复数字的序列,返回所有不重复的全排列. 示例: 输入: [1,1,2] 输出: [ [1,1,2], [1,2,1], [2,1,1] ] 分析:跟46题一样,只不过加了一些限制(包含了重复数字). AC代码(时间复杂度比较高,日后二刷的时候改进): class Solution { List<List<Integer>> ans = new ArrayList<>()…

目录 # 前端与算法 leetcode 350. 两个数组的交集 II 题目描述 概要 提示 解析 解法一:哈希表 解法二:双指针 解法三:暴力法 算法 # 前端与算法 leetcode 350. 两个数组的交集 II 题目描述 给定两个数组,编写一个函数来计算它们的交集. 示例 1: 输入: nums1 = [1,2,2,1], nums2 = [2,2] 输出: [2,2] 示例 2: 输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4] 输出: [4,9] 说明…

[LeetCode]95. Unique Binary Search Trees II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/unique-binary-search-trees-ii/description/ 题目描述: Given an integer n, generate all s…

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",…

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…

 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 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; ;…

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&…