寻找最大距离 [英] Find largest distance

问题描述

我遇到了一个问题，需要我想出一种方法来计算不同形状的最大边周长。我对如何开始使用它感到困惑。







这是一些代码。



公共类PerimeterAssignmentRunner {



public double getPerimeter（Shape s）{



double totalPerim = 0.0;



点prevPt = s.getLastPoint（）;



for（Point currPt：s.getPoints（））{



double currDist = prevPt.distance（currPt）;



totalPerim = totalPerim + currDist;



prevPt = currPt;



}



// totalPerim就是答案



return totalPerim;



}











包裹提供了另一个文件中给出的点数（不知道确切的术语）。



我尝试过：



double largestSide = 0 ;

点prevPt = s.getLastPoint（）;

for（Point currPt：s.getPoints（））

Hi, I am stuck on a problem that requires me to come up with a way to calculate the largest side perimeters of different shapes. I am kind of confused on how to even start with it tbh.



This is some of the code.

public class PerimeterAssignmentRunner {

public double getPerimeter (Shape s) {

double totalPerim = 0.0;

Point prevPt = s.getLastPoint();

for (Point currPt : s.getPoints()) {

double currDist = prevPt.distance(currPt);

totalPerim = totalPerim + currDist;

prevPt = currPt;

}

// totalPerim is the answer

return totalPerim;

}





The package was provided with points given in another file (don't know the exact the terminology).

What I have tried:

double largestSide = 0;
Point prevPt = s.getLastPoint();
for(Point currPt:s.getPoints())

推荐答案

计算每个部分的长度，即基本的毕达哥拉斯： BBC - GCSE Bitesize ：线段的长度 [ ^ ]

然后你所要做的就是锻炼时间最长......

Work out the length of each segment, that is basic Pythagoras: BBC - GCSE Bitesize: The length of a line segment[^]
Then all you have to do is work out the longest...

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