matrix

Get specific element from array in Java

由 旧巷老猫 提交于 2020-03-05 04:59:59
问题 Basically I am trying to return an element from an 2d array in java. I have created a separate Matrix class and inside the class I want to write a get_element method which would take as input the coordinates of the element I want from the matrix and the matrix itself, however I am not sure how to do this. public static double get_element(Matrix A, double m , double n) { for(int i=0;i<A.rows;i++) for(int j=0;j<A.cols;j++) return A.data[m][n]; } This is how my code look right now. And I get an

Get specific element from array in Java

由 ￣綄美尐妖づ 提交于 2020-03-05 04:59:07
问题 Basically I am trying to return an element from an 2d array in java. I have created a separate Matrix class and inside the class I want to write a get_element method which would take as input the coordinates of the element I want from the matrix and the matrix itself, however I am not sure how to do this. public static double get_element(Matrix A, double m , double n) { for(int i=0;i<A.rows;i++) for(int j=0;j<A.cols;j++) return A.data[m][n]; } This is how my code look right now. And I get an

Finding an element in a 2d array given its position on a spiral

由 一世执手 提交于 2020-03-05 02:10:33
问题 I am trying to solve a problem which involves a spiral ordering of the elements of a matrix and how to calculate the corresponding row and column. All the queries are in the form SZ P , where SZ is the size of the matrix and P the position in spiral starting from the center and ending in the top right corner. The output has to be the cartesian coordinates (row and column) of the P point in the spiral (starting with line 1 at the bottom and column 1 at the left). What I did to solve it, is

Finding an element in a 2d array given its position on a spiral

由 时光总嘲笑我的痴心妄想 提交于 2020-03-05 02:10:07
问题 I am trying to solve a problem which involves a spiral ordering of the elements of a matrix and how to calculate the corresponding row and column. All the queries are in the form SZ P , where SZ is the size of the matrix and P the position in spiral starting from the center and ending in the top right corner. The output has to be the cartesian coordinates (row and column) of the P point in the spiral (starting with line 1 at the bottom and column 1 at the left). What I did to solve it, is

查找二维数组中是否存在目标值

由 谁说胖子不能爱 提交于 2020-03-04 19:50:48
在一个二维数组中，该数组满足所有行从左至右递增，所有列满足从上到下递增。给定一目标值Target，若数组中存在该值，则返回true,否则返回false。 分析： 数字所有行从左到右递增，所有行从上到下递增，则每次将目标值与数组右上角的值进行比较，若相等则返回True,若小于该值,则到下一行，若大于该值，则到前一列。 代码如下： class Solution { public : bool findNumber ( vector < vector < int >> & matrix , int target ) { if ( matrix . empty ( ) ) return false ; int height = matrix . size ( ) - 1 ; int width = matrix [ 0 ] . size ( ) - 1 ; int h = 0 ; while ( h <= height && width >= 0 ) { if ( matrix [ h ] [ width ] == target ) return true ; else if ( matrix [ h ] [ width ] < target ) h ++ ; else width -- ; } return false ; } } 来源： CSDN 作者： 2020向前冲_ 链接：

Swift -How to update ARAnchor to follow Camera's position

由 霸气de小男生 提交于 2020-03-04 19:35:15
问题 I followed this code from @rickster which 100% works and looks great. In the video he's animating a SCNNode that is set using an ARAnchor from 1 position to another and back. I tried to do something similar except I want the node that is set with the ARAnchor to follow/update it's position to another node that is a child of the camera. I'm having a problem updating the position in func renderer(_ renderer: SCNSceneRenderer, willRenderScene scene: SCNScene, atTime time: TimeInterval) { } I

I need to alter my Matlab Gaussian elimination program to change the form of the final matrix before back substitution [closed]

由 时光怂恿深爱的人放手 提交于 2020-03-04 18:37:26
问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 15 days ago . I have made this Matlab function that performs Gaussian elimination however, I need to alter my code so that the final matrix (C) before back substitution is in the form of the linked picture named 'Required form of matrix'(where it says A = ). Right now my code performs Gaussian elimination

I need to alter my Matlab Gaussian elimination program to change the form of the final matrix before back substitution [closed]

由 微笑、不失礼 提交于 2020-03-04 18:36:51
问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 15 days ago . I have made this Matlab function that performs Gaussian elimination however, I need to alter my code so that the final matrix (C) before back substitution is in the form of the linked picture named 'Required form of matrix'(where it says A = ). Right now my code performs Gaussian elimination

How can I find the joint eigenvalues of two matrices in MATLAB?

由 十年热恋 提交于 2020-03-04 03:43:30
问题 If the joint eigenvalues of matrices A and B are defined as the roots of the equation det(lambda * A - B ) = 0, how can I solve this in MATLAB? In particular, I am not sure how exactly lambda is defined - it obviously needs to be a matrix or vector, as otherwise there would only be one joint eigenvalue. Also, I am not sure if there is any in-built function, or if, say, fzero for finding the roots of nonlinear functions needs to be used. 回答1: There is a built-in function for this. http://www

剑指offer---顺时针打印矩阵

由 点点圈 提交于 2020-03-03 19:04:20
题目描述 ：输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数 字，例如，如果输入如下4X4矩阵：12345678910111213141516则依次 打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10. 思路 ：按层模拟：终止行号大于起始行号，终止列号大于起始列号。 package Function ; import java . util . ArrayList ; import java . util . List ; public class spiralOrder29 { public List < Integer > spiralOrder ( int [ ] [ ] matrix ) { List < Integer > res = new ArrayList < > ( ) ; if ( matrix == null || matrix . length == 0 ) { return res ; } //行 int r1 = 0 , r2 = matrix . length - 1 ; //列 int c1 = 0 , c2 = matrix [ 0 ] . length - 1 ; while ( r1 <= r2 && c1 <= c2 ) { // 从 左 往 右 for ( int c = c1 ; c <=
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