[英]finding the index of largest and smallest numbers in a 2D Array
I'm really stumped trying to find the index of the largest and smallest numbers of a 5x5 array with random numbers up to 1000 generated into them. 我真的很难找到5x5阵列的最大和最小数字的索引,其中随机数量高达1000。 Here my code:
这是我的代码:
import java.util.Random;
public class MaxMinArray {
public static void main (String args[]) {
int x=0, y=0, max=0, min=1000;;
int[][] numbers = new int[5][5];
for (x=0; x<numbers.length; x++) { //outer for
for(y=0; y<numbers.length; y++) { //inner for
numbers[x][y]= (int)(Math.random()*1000); //random generator
if(max < numbers[x][y]) //max number
max = numbers[x][y];
if(min>numbers[x][y]) //min number
min = numbers[x][y];
int maxIndex = 0;
for(int index = 1; index<numbers.length; index++)
if(numbers[maxIndex]< numbers[index])
maxIndex = index;
}
}
System.out.println("Max number in array:" + max + " ");
System.out.println("Max number is in" + maxIndex + " ");
System.out.println("Min number in array:" + min + " ");
}
}
You should keep track of both the x
and y
index of the maximum/minimum element. 您应该跟踪最大/最小元素的
x
和y
索引。 No need to post-process, it's just a matter of bookkeeping: 无需后期处理,只需记账:
if(max < numbers[x][y]) {
max = numbers[x][y];
maxX = x;
maxY = y;
}
Use a Point
to keep track of your indices. 使用
Point
来跟踪您的指数。
Point min = new Point(0, 0);
Point max = new Point(0, 0);
for(int[] row: numbers) {
for(int col= 0; col < row.length; col++) {
if(numbers[row][col] < numbers[min.X][min.Y])
{max.X = row; min.Y = col;}
if(numbers[row][col] > numbers[max.X][max.Y])
{max.X = row; max.Y = col;}
}
}
if(numbers.length > 0) {
System.out.println(numbers[min.X][min.Y] + " is the minimum.");
System.out.println(numbers[max.X][max.Y] + " is the maximum.");
}
for something of this small of scale, a simple double for loop should be the simplest to understand and utilize. 对于这种小规模的东西,简单的双循环应该是最容易理解和利用的。
int n=5;
int min = array[0][0];
int[] minIndex = {0,0};
int max = array[0][0];
int[] maxIndex = {0,0};
for (int i=0; i<n; i++)
{
for (int j=0; j<n; j++)
{
if (array[i][j] < min)
{
min = array[i][j];
minIndex[0] = i;
minIndex[1] = j;
}
if (array[i][j] > max) {
max = array[i][j];
maxIndex[0] = i;
maxIndex[1] = j;
}
}
}
For non-trivial dimensions this might be a slow method, but for this size matrix n^2 complexity is fine. 对于非平凡的维度,这可能是一种缓慢的方法,但对于这个大小矩阵,n ^ 2复杂度很好。
EDIT: WOW, I missed the part about the indices. 编辑:哇,我错过了关于指数的部分。
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