J Hello word】的更多相关文章

jave 学习 public class Hello { public static void main (string args []) { int i = 0; for (i = 0; i < 3 ; i++ ) { system.out.println("Hello word!"); } } } 编译方法 javac hello.java 运行 java hello 规范:java 编程中类的名字要大写,而文件的名字要和类名保持一致 java 的数据类型 基本数据类型,和引…
Given a non-empty string s and an abbreviation abbr, return whether the string matches with the given abbreviation. A string such as "word" contains only the following valid abbreviations: ["word", "1ord", "w1rd", &…
使用的别人的思路,用一个二维数组记录每一个位置是否用过,然后通过递归来判断每一个位置是否符合 public class Solution { public boolean exist(char[][] board, String word) { if(board.length == 0) return false; int leni = board.length; int lenj = board[0].length; boolean[][] isVisited = new boolean[le…
Given a non-empty string s and an abbreviation abbr, return whether the string matches with the given abbreviation. A string such as "word" contains only the following valid abbreviations: ["word", "1ord", "w1rd", &…
Word Search I 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…
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…
Problem Description: http://oj.leetcode.com/problems/word-search/ Basic idea: recursively go forward, use char '#' to mark the char visited, so no extra memory is need, don't forget to recover the previous char if the program cannot go forward. class…
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 Description In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find the letters that appear in the answer it is necessary to unscramble four wor…
C# 处理Word自动生成报告 一.概述 C# 处理Word自动生成报告 二.数据源例子 C# 处理Word自动生成报告 三.设计模板 C# 处理Word自动生成报告 四.程序处理 现在说一下程序处理部分,有点长 本来是想做针对doc和docx的模板两个版本, 后来想到可以在生成的时候saveas里设置格式, 所以此版只支持对docx的模板处理, 想要doc的情况可以选择生成格式为doc的. 上代码: public class WordHelper { private Word.Applicat…
发现字典里面的单词数目多且长度短,可以用字典树保存 f[i]表示s[i~L]的分割方式,则有f[i]=∑f[i+len(word[j])]   其中word[j]为s[i~L]的前缀 注意字典树又叫前缀树,所以用前缀更方便,否则按顺序dp的话就要把字符倒序了 复杂度O(L*l) L为字符串长度,l为单词长度 #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #de…
将很早之前写的一个小组件重新整理优化一下,做成一个通用的功能.适用于导出数据库的结构(表.字段等)到Word或将体检数据自动生成Word版的体检报告等.代码:Github 一.主要需要完成功能: 1. 灵活的配置规则及word样式设置(文本.表格.图表.颜色等). 2. 支持表格. 3. 支持图表. 4. 支持章节内容循环生成. 5. 支持目录. 6.支持文档结构图 7.更新指定位置的文字 8.支持pdf导出. 最后结果如下: 图一 图二 图三 二.需求分析与实现方式   功能主要涉及3个比较重…
Win:显示开始菜单 Win + E:打开文件管理器 Win + D:显示桌面 Win + L:锁定计算机 Win + I:打开设置 Win + M:最小化所有窗口 Alt + F4:1.用来关闭当前窗口:2.如所有窗口已关闭或最小化,则为关机快捷键,使用后会弹出一关机对话框. Alt + Tab:在打开的应用之间进行切换 Alt + ←:1.资源管理器中,查看上一级文件夹:2.浏览器中,返回至上一次访问的历史记录页面 Alt + →:1.资源管理器中,查看下一级文件夹:2.浏览器中,跳转至下一…
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…
  import os import os.path from win32com import client as wc c=[] rootdir=["d:/77"] #以该路径为实验 def txt(j,c): word = wc.Dispatch('Word.Application') doc = word.Documents.Open(c[j]) newname=c[j][:-5]+"(translate txt)" doc.SaveAs(newname,4)…
2018-10-26 00:32:05 问题描述: 问题求解: 方法一.Trie 最长出现的字符串,最容易想到的解法就是Trie树了,于是首先使用Trie树进行了实现,代码量有点大,当然了是可以A掉的,只是对于这种Easy的题,理论上是不该超过50行代码的. public class MostCommonWord { class TrieNode { public TrieNode[] next = new TrieNode[26]; public int cnt = 0; public Str…
先写上我的代码: 我总是不知道何时把任务交给下一个递归.以致于,写出的代码很臃肿! 放上别人递归的简洁代码: bool exist(char** board, int m, int n, char* word) { if(word == NULL) return true; ) return false; bool ans= false; bool **used = (bool**)malloc(m*sizeof(bool*)); ; i < m; i++) { used[i] = (bool*…
首先将word模板中需要填写excel中数据的空白处用自己独特的字符串标记,比如   数据001  什么的.如下图: 这样,就可以用vba搜寻这些自己独特的标记来根据excel内容填充word了. 第一个填好的如下图: 而excel中数据如下图: 然后代码如下: Sub 生成Word文件_Click() Dim Word对象 As New Word.Application, 当前路径, 导出文件名, 导出路径文件名, i, j Dim Word表格行号, Word表格列号, Excel数据表列号…
题目链接:https://leetcode.com/problems/word-search/#/description 给出一个二维字符表,并给出一个String类型的单词,查找该单词是否出现在该二维字符表中. For example,Given board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] word = "ABCCED", -> returns true,word = "SEE…
Word SearchGiven 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 ma…
[抄题]: 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…
终于搞定了这个DFS,最近这个DFS写的很不顺手,我一直以为递归这种东西只是在解重构时比较麻烦,现在看来,连最简单的返回true和false的逻辑关系都不能说one hundred present 搞定. 人品啊TLE: class Solution { public: bool legal(int i, int j, vector<vector<char>> board) { && i < board.size() && j >= &a…
题目: 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 b…
题目 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…
本题有几个注意点: 1. 回溯找路径时.依据路径的最大长度控制回溯深度 2. BFS时,在找到end单词后,给当前层做标记find=true,遍历完当前层后结束.不须要遍历下一层了. 3. 能够将字典中的单词删除.替代visited的set,这样优化以后时间从1700ms+降到800ms+ 代码例如以下: class Solution { public: vector<vector<string>> findLadders(string start, string end, uno…
题目: Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word must exist in the wor…
题目: Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: Only one letter can be changed at a time Each transformed word must exist in the word list.…
------------------------------------------------------------第七周:Follow up question 1,寻找峰值 寻找峰值 描述 笔记 数据 评测 你给出一个整数数组(size为n),其具有以下特点: 相邻位置的数字是不同的 A[0] < A[1] 并且 A[n - 2] > A[n - 1] 假定P是峰值的位置则满足A[P] > A[P-1]且A[P] > A[P+1],返回数组中任意一个峰值的位置. 注意事项 数…
#include <math.h> #include <time.h> #include <stdlib.h> #include <iostream> #include <windows.h> using namespace std; #define N 1517 //文献数目 #define textN 6 //题目数目 #define ERROR 1 #define OK 0 const int WORD_LENGTH = 30;//定义单个…
一.前言 在之前写过一个词频统计的C语言课设,别人说你一个大三的怎么写C语言课程,我只想说我是先学习VB,VB是我编程语言的开始,然后接触到C语言及C++:再后来我是学习C++,然后反过来学习C语言,记得那时候自己在C++中没有好好学习,考试之前疯狂的背代码,然后过了.后来学习C语言的时候,自己知道基础很重要,然后认真学习.这WC这回自己就轻车熟路,记得那时候丁国辉课设老师,问我有多少是自己写的,我说有80%自己写的,然后他让我当场给程序增加一个总的单词数和每一个单词的频率,当时记得自己在旁边改…