问题
I know how to go through whole array, but I only need number of duplicate occurrences. I'm at beginners level, so just basic use of loops and arrays.
int[] array = {12, 23, -22, 0, 43, 545, -4, -55, 43, 12, 0, -999, -87};
for (int i = 0; i < array.length; i++) {
int count = 0;
for (int j = 0; j < array.length; j++) {
count++;
}
System.out.println(array[i] + "\toccurs\t" + count + "X");
}
回答1:
You can do better if you use more than just loops and arrays, but a simple algorithm would be to use two nested for
loops, and put an if
statement inside that increments a counter when a duplicate is found.
int[] array = {12, 23, -22, 0, 43, 545, -4, -55, 43, 12, 0, -999, -87};
for (int i = 0; i < array.length - 1; i++) {
int count = 1;
for (int j = i + 1; j < array.length; j++) {
if (array[i] == array[j]) {
count++;
}
}
if (count > 1) {
System.out.println(array[i] + "\toccurs\t" + count + " times");
}
}
回答2:
using System;
public class Exercise34
{
public static void Main()
{
int[] a = { 3,4,5,6,7,8,3,4,5,6,7,8,9,9};
int n = a.Length-1;
int dupcounter = 0;
for (int i = 0; i < n; i++)
{
int counter = 0;
for (int j = i + 1; j <= n; j++)
{
if (a[i] == a[j])
{
counter++;
n--;
if (counter == 1)
{
dupcounter++;
Console.WriteLine(a[i]);
}
for (int k = j; k <= n; k++)
{
a[k] = a[k + 1];
}
}
}
}
Console.WriteLine(dupcounter);
}
}
来源:https://stackoverflow.com/questions/40685936/how-to-count-and-print-out-only-duplicates