Z字形编排问题详解(C++): 问题描述:给定一个矩阵matrix,输出矩阵matrix进行Z字形编排后的内容. 原矩阵: 输出形式: 算法分析与详细解答: 要解决这样一个问题,可能一开始无从下手,但是我们只要认真观察Z字形矩阵的走向过程,就不难发现其中的规律.对于原始矩阵matrix中的任意元素 matrix[i][j]的遍历走向规律可以分为以下三种情况: 1.如果矩阵的元素matrix[i][j]中纵坐标j是偶数,并且i==0或i==SIZE-1,那么遍历路径在矩阵中的走向就是向右移动一格.

/* * BitSets are packed into arrays of "words." Currently a word is * a long, which consists of 64 bits, requiring 6 address bits. * The choice of word size is determined purely by performance concerns. */ private final static int ADDRESS_BITS_P

Given an integer matrix, find the length of the longest increasing path. From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed). E

1.Implement an algorithm to determine if a string has all unique characters What if you can not use additional data structures? The length of ACSII code of a character is 8, so we can build a array, the length is 260, to represent the hash table of a

给定一个二维矩阵,计算其子矩形范围内元素的总和,该子矩阵的左上角为 (row1, col1) ,右下角为 (row2, col2). 上图子矩阵左上角 (row1, col1) = (2, 1) ,右下角(row2, col2) = (4, 3),该子矩形内元素的总和为8.示例:给定 matrix = [  [3, 0, 1, 4, 2],  [5, 6, 3, 2, 1],  [1, 2, 0, 1, 5],  [4, 1, 0, 1, 7],  [1, 0, 3, 0, 5]]sumRegi

题目要求: 给定一个整数矩阵,找出最长递增路径的长度. 对于每个单元格,你可以往上,下,左,右四个方向移动. 你不能在对角线方向上移动或移动到边界外(即不允许环绕). 示例: 输入: nums = [ [9,9,4], [6,6,8], [2,1,1]] 输出: 4 解释: 最长递增路径为 [1, 2, 6, 9]. class Solution { public: int dx[5] = {-1, 0, 1, 0}; int dy[5] = {0, 1, 0, -1}; int longest

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. For example, given the array [2,3,1,2,4,3] and s = 7,the subarray [4,3] has the minimal

Problem: Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. For example, given the array [2,3,1,2,4,3] and s = 7,the subarray [4,3] has the

Java中length,length(),size()区别 length属性:用于获取数组长度. eg: int ar[] = new int{1,2,3} /** * 数组用length属性取得长度 */ int lenAr = ar.length;//此处lenAr=3 System.out.println("Arr length:"+lenAr); length()方法:用于获取字符串长度. String str = "Hello World Java"; /

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. For example, given the array [2,3,1,2,4,3] and s = 7,the subarray [4,3] has the minimal

length属性:用于获取数组长度. eg: int ar[] = new int{1,2,3} /** * 数组用length属性取得长度 */ int lenAr = ar.length;//此处lenAr=3 System.out.println("Arr length:"+lenAr); length()方法:用于获取字符串长度. String str = "Hello World Java"; /** * 字符串用length()方法取得长度 */ int

别人的代码 class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { int l, r, cum, res = nums.size()+1; l = r = cum = 0; while ((unsigned int)r < nums.size()) { cum += nums[r++]; while (cum >= s) { res = min(res, r-l); cum -= num

/** * Created by leo on 16/4/30. */ public interface GanchaiService { @GET("digest?t={type}&p={page}&size={count}") Call<List<GanChaiEntry>> ListGanchaiEntry(@Path("type") int type , @Path("count") int cou

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. For example, given the array [2,3,1,2,4,3] and s = 7,the subarray [4,3] has the minimal

线性表的顺序存储结构:是指用一组地址连续的存储单元一次存放线性表的元素.为了使用顺序结构实现线性表,程序通常会采用数组来保存线性中的元素,是一种随机存储的数据结构,适合随机访问.java中ArrayList类是线性表的数组实现. 下面是自己模仿ArrayList实现的最简单的线性表的顺序存储功能 import java.util.Arrays; //线性表的顺序存储 抽象类的实现public class SequenceList<T> {    private int DEFAULT_SIZE

项目需要,根据所输入的字符串,的需要计算串帐户Size. 包代码如下面.只需要传递一个字符串,您可以返回Size: Size ChartDemoScene::calculateFontSize(const char *str ) { std::string tempString = str; log("tempString = %s",tempString.c_str()); size_t computeCount = tempString.size(); //假设字符串非常长每次抽取

传送门 题意: 一棵树,询问子树中权值大于$k$的节点个数,修改点权值,插入新点:强制在线 一开始以为询问多少种不同的权值,那道CF的强制在线带修改版,直接吓哭 然后发现看错了这不一道树上分块水题... 用王室联邦分块的话需要维护每一个块$dfs$序最小值和最大值,并且插入操作会破坏原来的性质 不如直接按$size$分块,根节点$size<block$就加入根,否则新建块 $size$分块不能保证块的数量,可以被菊花图卡掉,然而本题没有所以就可以安心的写了 每个块维护排序后的值 查询操作,不完整

An integer interval [a, b] (for integers a < b) is a set of all consecutive integers from a to b, including a and b. Find the minimum size of a set S such that for every integer interval A in intervals, the intersection of S with A has size at least

最近在看HashMap的源码,有很多概念都很模糊,今天写了一个测试例子,加深对这几个概念的理解,并演示了扩容及树化的过程(见下篇博文:). 注:本文基于JDK 1.8  HashMap的结构 约定 约定前面的数组结构的每一个格格称为桶 约定桶后面存放的每一个数据称为bin bin这个术语来自于JDK 1.8的HashMap注释. size size表示HashMap中存放KV的数量(为链表和树中的KV的总和). capacity capacity译为容量.capacity就是指HashMap中桶

class Solution { public: vector<vector<int>> threeSum(vector<int>& a) { vector<vector<int>> ans; sort(a.begin(),a.end()); int n = a.size(); //for(int k = 0; k < a.size() -2; k++) ; k < n -; k++) //问题在这里.a.size()是uns

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead. Example: Input: [2,3,1,2,4,3], s = 7 Output: 2 Explanation: the subarray [4,3