我的问题是当我们将j=i+1
更改为j = 0
或更改时会发生什么
j = i
。我的意思是这两种情况有什么区别?
import java.util.Scanner;
public class FindNearestPoints {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter the number of points: ");
int numberOfPoints = input.nextInt();
// Create an array to store points
double[][] points = new double[numberOfPoints][2];
System.out.print("Enter " + numberOfPoints + " points: ");
for (int i = 0; i < points.length; i++) {
points[i][0] = input.nextDouble();
points[i][1] = input.nextDouble();
}
// p1 and p2 are the indices in the points array
int p1 = 0, p2 = 1; // Initial two points
double shortestDistance = distance(points[p1][0], points[p1][1],
points[p2][0], points[p2][1]); // Initialize shortestDistance
// Compute distance for every two points
for (int i = 0; i < points.length; i++) {
for (int j = i + 1; j < points.length; j++) {
double distance = distance(points[i][0], points[i][1],
points[j][0], points[j][1]); // Find distance
if (shortestDistance > distance) {
p1 = i; // Update p1
p2 = j; // Update p2
shortestDistance = distance; // Update shortestDistance
}
}
}
// Display result
System.out.println("The closest two points are " +
"(" + points[p1][0] + ", " + points[p1][1] + ") and (" +
points[p2][0] + ", " + points[p2][1] + ")");
}
/** Compute the distance between two points (x1, y1) and (x2, y2)*/
public static double distance(
double x1, double y1, double x2, double y2) {
return Math.sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}
}
答案 0 :(得分:2)
如果设置j = i
,那么最小距离将为零,因为您要计算点与自身之间的距离!
如果您设置了j = 0
,则上述内容仍适用,但由于您要包含j == i
,因此会包含不必要的重复内容。
答案 1 :(得分:-1)
当我们将mEditText.setOnKeyListener((v, keyCode, event) -> {
//for submiting result
if ((keyCode == KeyEvent.KEYCODE_SPACE || keyCode == KeyEvent.KEYCODE_ENTER) &&
mEditText.getText().toString().length() > 0) {
// Your Positive button's code over here
}
return false;
});
更改为j=i+1
或j = 0
我们第一次到达代码中的j = i
时,会生成变量j并初始化为for statement
。
将其更改为任何其他值,我们将获得不同的初始化值。