On a N * N grid, we place some 1 * 1 * 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Return the total surface area of the resulting shapes. Example 1: Input: [[2]] Output: 10 Example 2: Input: [[1…
题目要求 On a N * N grid, we place some 1 * 1 * 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Return the total surface area of the resulting shapes. 题目分析及思路 给定一个N*N网格,在其上放置1*1*1的小方块,二维数组的值即为当前位置的高度,要求…
problem 892. Surface Area of 3D Shapes 题意:感觉不清楚立方体是如何堆积的,所以也不清楚立方体之间是如何combine的.. Essentially, compute the surface area of each grid but substract the overlapping areas. All areas = surface + combined area so we have surface = * total_count - * combi…
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 日期 题目地址:https://leetcode.com/problems/surface-area-of-3d-shapes/description/ 题目描述 On a N * N grid, we place some 1 * 1 * 1 cubes. Each value v = grid[i][j] represents a tower o…
问题 NxN个格子中,用1x1x1的立方体堆叠,grid[i][j]表示坐标格上堆叠的立方体个数,求这个3D多边形的表面积. Input: [[1,2],[3,4]] Output: 34 思路 只要把每个柱体的表面积加起来(grid[i][j] * 4 ,4表示四个侧面,2表示上下两个面),然后减去重叠的部分即可. 重叠的部分为x方向(或y方向)上相邻柱体中较小的grid值. 时间复杂度O(n^2),空间复杂度O(1) 代码 class Solution(object): def surfac…
On a N * N grid, we place some 1 * 1 * 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Return the total surface area of the resulting shapes.   Example 1: Input: [[2]] Output: 10 Example 2: Input: […
题目如下: 解题思路:对于v = grid[i][j],其表面积为s = 2 + v*4 .接下来只要在判断其相邻四个方向有没有放置立方体,有的话减去重合的面积即可. 代码如下: class Solution(object): def surfaceArea(self, grid): """ :type grid: List[List[int]] :rtype: int """ res = 0 for i in range(len(grid)):…
On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, and z axes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Now we view the projection of these cubes onto the xy, yz, and…
问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/4136 访问. 在 N * N 的网格上,我们放置一些 1 * 1 * 1  的立方体. 每个值 v = grid[i][j] 表示 v 个正方体叠放在单元格 (i, j) 上. 返回结果形体的总表面积. 输入:[[2]] 输出:10 示例 2: 输入:[[1,2],[3,4]] 输出:34 示例 3: 输入:[[1,0],[0,2]] 输出:16 示例 4: 输…
On a N * N grid, we place some 1 * 1 * 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Return the total surface area of the resulting shapes. Example 1: Input: [[2]] Output: 10 Example 2: Input: [[1…
题目要求 On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, and z axes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Now we view the projection of these cubes onto the xy, yz,…
在 N * N 的网格上,我们放置一些 1 * 1 * 1  的立方体. 每个值 v = grid[i][j] 表示 v 个正方体叠放在单元格 (i, j) 上. 返回结果形体的总表面积. 示例 1: 输入:[[2]] 输出:10 示例 2: 输入:[[1,2],[3,4]] 输出:34 示例 3: 输入:[[1,0],[0,2]] 输出:16 示例 4: 输入:[[1,1,1],[1,0,1],[1,1,1]] 输出:32 示例 5: 输入:[[2,2,2],[2,1,2],[2,2,2]]…
问题 NxN个格子中,用1x1x1的立方体堆叠,grid[i][j]表示坐标格上堆叠的立方体个数,求三视图面积. Input: [[1,2],[3,4]] Output: 17 Explanation: 见下图 思路 对于俯视图,只要一个格子有值,面积值就加1. 对于正视图(面朝x轴),对于某一个x,在y轴方向上拥有的最高grid值,表示,该x顺着y轴看过去看到的面积值. 对于侧视图(面朝y轴),对于某一个y,在x轴方向上拥有的最高grid值,表示,该y顺着y轴看过去看到的面积值. 把这些面积值…
problem 883. Projection Area of 3D Shapes 参考 1. Leetcode_easy_883. Projection Area of 3D Shapes; 完…
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 数学计算 日期 题目地址:https://leetcode.com/problems/projection-area-of-3d-shapes/description/ 题目描述 On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, an…
On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, and z axes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Now we view the projection of these cubes onto the xy, yz, and…
题目如下: 解题思路:分别求出所有立方体的个数,各行的最大值之和,各列的最大值之和.三者相加即为答案. 代码如下: class Solution(object): def projectionArea(self, grid): """ :type grid: List[List[int]] :rtype: int """ front = [0] * len(grid) side = [0] * len(grid) top = 0 for i in…
On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, and z axes. Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j). Now we view the projection of these cubes onto the xy, yz, and…
在 N * N 的网格中,我们放置了一些与 x,y,z 三轴对齐的 1 * 1 * 1 立方体. 每个值 v = grid[i][j] 表示 v 个正方体叠放在单元格 (i, j) 上. 现在,我们查看这些立方体在 xy.yz 和 zx 平面上的投影. 投影就像影子,将三维形体映射到一个二维平面上. 在这里,从顶部.前面和侧面看立方体时,我们会看到"影子". 返回所有三个投影的总面积. 示例 1: 输入:[[2]] 输出:5 示例 2: 输入:[[1,2],[3,4]] 输出:17 解…
  - BVH with SAH (Bounding Volume Hierarchy  with Surface Area Heuristic) -      0. Overview 包围层次盒(BVH)是一种基于"物体“(object, 区别于“空间”,spatial )的场景管理技术,广泛应用于碰撞检测.射线相交测试之类的应用场合中. 一般情况下BVH的性能都比较不错,起码比基于格子(Grid)的场景管理技术要高.   BVH的数据结构其实就是一棵二叉树(Binary Tree).它有两种…
here were various changes to memory related DMVs, DBCC memory status, and Perfmon counters in SQL Server 2012 as part of the redesign of the Memory Manager component of SQLOS. The Memory Manager redesign resulted in being able to more accurately size…
此主要讨论图像处理与分析.虽然计算机视觉部分的有些内容比如特 征提取等也可以归结到图像分析中来,但鉴于它们与计算机视觉的紧密联系,以 及它们的出处,没有把它们纳入到图像处理与分析中来.同样,这里面也有一些 也可以划归到计算机视觉中去.这都不重要,只要知道有这么个方法,能为自己 所用,或者从中得到灵感,这就够了. 注意:Registration可翻译为“配准”或“匹配”,一般是图像配准,特征匹配(特征点匹配). 15. Image Registration图像配准最早的应用在医学图像上,在图像融合…
Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Assume that the total area is never beyond the maximum possible value of int. 这道…
Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Find the maximum area of an island…
You have a list of points in the plane. Return the area of the largest triangle that can be formed by any 3 of the points. Example: Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]] Output: 2 Explanation: The five points are show in the figure below. T…
题目标签:HashMap 题目给了我们一组 xy 上的点坐标,让我们找出 能组成矩形里最小面积的那个. 首先遍历所有的点,把x 坐标当作key 存入map, 把重复的y坐标 组成set,当作value 存入map. 然后遍历所有的点,找出 对角的两个点, 再去map里确认是否存在剩下的两个对角点,计算面积,一直保留最小的那个面积. Java Solution: Runtime: 214 ms, faster than 78.80% Memory Usage: 61.5 MB, less than…
We are given a list of (axis-aligned) rectangles.  Each rectangle[i] = [x1, y1, x2, y2] , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the ith rectangle. Find the total area…
Find the total area covered by two rectilinearrectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Example: Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2 Output:…
给定一些矩形2 求覆盖面积 矩形不超过200个 1 算法1 朴素思想 虽然朴素但是代码却有意思 利用容斥原理 复杂度高达 N*2^N class Solution: def intersect(rec1,rec2): return [max(rec1[1],rec2[1]), max(rec1[2],rec2[2]), min(rec1[3],rec2[3]), min(rec1[4],rec2[4] ] #这里是两个矩形的两点式求 矩形的相交子矩形 *非常值得思考 def area(rec):…
Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Assume that the total area is never beyond the maximum possible value of int. Cr…