日常项目开发中，一般流程是产品经理提出需求，相关人员进行需求评审，然后是前后端工程师开发功能，再到测试、发布上线。 流程如下： 图-1 可以看到，这些步骤是存在先后（依赖）关系的：前
前一段时间调试一个程序的，半天没发现问题在哪里，经过测试才发现 delete adjacent duplicates from itab 和 delete adjacent duplicates from itab comparing all fields还是有区别的： 前者相邻两行数据，如果除金额字段以外的其他字段都相同，则去重复删除其中一行； 后者相邻两
原题链接在这里：https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/ 题目： There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums. You a
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tol=0; adjacent_point_dist=[]; % 这个变量用来存储车道线上相邻点之间的距离 points_veryclose=[]; %2021.12.25 通过定义 adjacent_point_dist这个变量发现，相邻点之间的距离个别是0.2mm，这个不正常，这样相邻的点应该去掉其中一个 for k=2:size(Thefirstline,1) tol=t
// 问题 G: 4-adjacent 时间限制: 1.000 Sec 内存限制: 128 MB 题目描述 We have a sequence of length N, a=(a1,a2,…,aN). Each ai is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: For each 1
重点单词： broadcast v.散布，传播，广播， n.电视节目，广播节目； allocate v.分配， 分派， 划拨。 singular adj.单数的，突出的，独特的。 format n.计划， 设计， 安排 v.格式化， 排版 segment v.划分， 分割, 分裂；n.部分 片段。 symmetric adj.对称的. retransmit v.重新发送， 中继 ，转播；upper case n.
这道题很简单，其实是一个Stack的问题， 但是不用Stack，用一个StringBuilder就可以解决，时间复杂度和空间复杂度都是O(n). public String removeDuplicates(String s) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < s.length(); i++) {
SORT <内表> BY <字段> [ascending/descending]. DELETE ADJACENT DUPLICATES FROM <内表> COMPARING <字段> *删除所有字段相同数据SORT <内表> BY <字段> <字段>[ascending/descending]. DELETE ADJACENT DUPLICATES FROM <内表> COMPARING ALL FIEL
这道题目如果是暴力肯定不行，标签是入门，看着是个数学题。 如何判断两个数的积是 \(4\) 的倍数呢？ 两个数中其中一个数含有因数 \(4\) 两个数的积中含有因数 \(4\)，即两个数都是 \(2\) 的倍数（但这两个数都不含因数 \(4\)） 所以我们只要边读入边把 \(4\) 的倍数和是 \(2\) 的倍数但不
题目 分析 首先发现大的字母如果存在对于小的字母的删除来说一定不优，于是想到可以直接枚举字母 \(z\) 到 \(a\) 依次删除。 删除的过程可以用链表模拟也可以直接 \(vector\) 暴力。 代码 #include<bits/stdc++.h> using namespace std; //#ifdef ONLINE_JUDGE // #define getchar
题目来源：http://poj.org/problem?id=1804 Background Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just by glancing at them. And he can even count Poker cards. Charlie would lov
前言
adjacent_find用于在指定范围内查找1个连续出现2次的元素。 #include <algorithm> #include <iostream> #include <string> #include <vector> using namespace std; int main() { vector<int> v{ 1,2,2,4,5,2,3,4,7,8,9,1,7,7,3,4 }; auto it = adjacen
将\(1 \sim n^2\)放置在\(n\times n\)的棋盘当中，要求当前格子中的数字与其四连通的格子中的数字不相邻。 考虑将当前格子染成黑色，与其四连通的格子染成白色，即下标\((i,j)\)的格子，当\(i+j\)为偶数时染成黑色，\(i+j\)为奇数时染成白色，则颜色不同的格子中的数字不能相邻。 按从左至右
We will consider the numbers
There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums. You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i]
Can Place Flowers (E) 题目 You have a long flowerbed in which some of the plots are planted, and some are not. However, flowers cannot be planted in adjacent plots. Given an integer array flowerbed containing 0's and 1's, where 0 means empty and 1
Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them. We repeatedly make duplicate removals on S until we no longer can. Return the final string after all such duplicate removal
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
题目链接 2s+n=100，可以直接暴力贪心，从末尾z开始删到a即可 #include<bits/stdc++.h> using namespace std; #define ms(x,y) memset(x, y, sizeof(x)) #define lowbit(x) ((x)&(-x)) typedef long long LL; typedef pair<int,int> pii; void run_case() { int n; string
链接：https://codeforces.ml/contest/1321/problem/C You are given a string ss consisting of lowercase Latin letters. Let the length of ss be |s||s|. You may perform several operations on this string. In one operation, you can choose some index ii and
FIT3143 – Assignment 2Marks: 20 marks and 20% of all marks for the unit.Due Date: Week 11, Friday 18/October/2019, 11:55 PM Melbourne time.Submission: (i) The assignment submission must be made through Moodle for this unit by thedue date. (ii) The submiss
adjacent_difference()是位于numeric头文件中，用来比较序列中相邻元素的差值，并将它们的差值保存到另外一个序列上。 因为第一个的前一个值为空，所以第一个元素会原封不动的复制到序列中，然后第二个元素和第一个元素相减结果作为新的第二个序列元素，再用第三个元素减去第二个元素
查找算法——adjacent_find 功能描述： 查找相邻重复元素 函数原型： adjacent_find(iterator beg, iterator end); // 查找相邻重复元素,返回相邻元素的第一个位置的迭代器 // beg 开始迭代器 // end 结束迭代器 测试代码 #include <iostream> using namespace std; #i
