我正在尝试计算数组中存在唯一数字的次数，使用的索引数量取决于输入的元素数量。主要是操作除了1.第一个值没有被考虑。循环正在检查arrays.length -1所以数字3显示为1的计数，即使我输入3两次。我知道修复此问题的最佳方法是运行一个不使用arrays.length -1的循环，但是我无法将条目与旁边的条目进行比较，如if(a[i] == a[i + 1] && a[i] != 0)，以查看是否存在超过一次出现的值。我认为最好的办法是将计数值存储在我的count方法中，并计算相应的数组值，然后在方法之外做一个for循环可能吗？我无法看到这样做的方式，因为我对java很新。我可以提供一些指导：）
import java.util.Scanner;
public class Prac_ExeOne
{
static int count = 1;
static int numberUsed = 0; // static its used to show its a class wide variable and there is only one copy.
static int[] Array = new int [50]; // the maximum elements inside the array that can be used is 10;
int size;
public int fillArray(int[] a)
{
System.out.println("Enter up to " + a.length + " nonnegative numbers.");
System.out.println("Mark the end of the list with a negative number.");
Scanner keyboard = new Scanner(System.in);
int next;
int index = 0;
next = keyboard.nextInt();
while ((next >= 0) && (index < a.length ))
{
numberUsed++;
a[index] = next;
index++;
// Print out each value of next
System.out.println(next);
next = keyboard.nextInt();
//System.out.println("Number of indexes used" + numberUsed);
}
keyboard.close(); // close the keyboard so it can't continue to be used.
return index;
}
public int[] sort(int[] arrays)
{
for(int i = 0;i < arrays.length -1 ;i++ )
{
int store = 0;
// Move Larger Values to the right.
if (arrays[i + 1 ] < arrays[i])
{
store = arrays[i];
arrays[i] = arrays[i + 1];
arrays[i + 1] = store;
}
// Sort swapped smaller values to the left.
for(int j = i; j > 1; j--)
{
if (arrays[j] < arrays[j - 1])
{
store = arrays[j];
arrays[j] = arrays[j - 1];
arrays[j - 1] = store;
}
}
}
return arrays;
}
public void count(int[] a)
{
//for each element in array go through if conditons.
System.out.println("N " + "Count");
for(int i = 0;i < a.length -1;i++)
{
if(a[i] == a[i + 1] && a[i] != 0)
{
count++;
}
if(a[i] != a[i+1])
{
count = 1;
}
if (a[i] != 0)
{
System.out.println(a[i] + " " + count);
}
}
}
public static void main(String[] args)
{
Prac_ExeOne score = new Prac_ExeOne();
score.fillArray(Array);
score.sort(Array);
score.count(Array);
}
}
输入：
Enter up to 50 nonnegative numbers.
Mark the end of the list with a negative number.
3
3
2
2
-2
输出：
N Count
3 1
2 2
2 1
期望的结果：
简而言之，我希望程序正确计算值，然后在N下显示左边的值，并在数组中显示它在Count右下方的次数
答案 0 :(得分：1)
简而言之，我希望程序正确计算值
计入Map：
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class App {
public static void main(String[] args) throws Exception {
List<Integer> ints = new ArrayList<>();
ints.add(3);
ints.add(3);
ints.add(2);
ints.add(2);
ints.add(-2);
ints.add(5);
Map<Integer, Integer> counts = new HashMap<>();
for (Integer i : ints) {
if (i < 0) {
break;
}
if (counts.containsKey(i)) {
counts.put(i, counts.get(i) + 1);
} else {
counts.put(i, 1);
}
}
System.out.println(counts);
}
}
输出：
{2=2, 3=2}
答案 1 :(得分：1)
适用于sort功能，
for (int j = i; j > 1; j--)
应该是
for (int j = i; j > 0; j--)
我在调用之后放System.out.println(Arrays.toString(Array));并在开头看到3，这导致我跳过了第一个元素。
请注意，还有way more efficient sorting algorithms。
对于count功能，您在错误的时间重置count并经常打印。我改变如下：
public void count(int[] a)
{
//for each element in array go through if conditions.
System.out.println("N " + "Count");
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] != 0)
{
if (a[i] == a[i + 1])
{
count++;
}
// if the next element is different, we already counted all of the
// current element, so print it, then reset the count
else
{
System.out.println(a[i] + " " + count);
count = 1;
}
}
}
// we haven't processed the last element yet, so do that
if (a[a.length-1] != 0)
System.out.println(a[a.length-1] + " " + count);
}
答案 2 :(得分：1)
如果你真的想使用带有唯一计数器的数组，你可以使用下面的代码：
public class Prac_ExeOne {
static int count = 1;
static int numberUsed = 0; // static its used to show its a class wide variable and there is only one copy.
static Integer[] Array = new Integer[50]; // the maximum elements inside the array that can be used is 10;
int size;
public int fillArray(Integer[] a) {
System.out.println("Enter up to " + a.length + " nonnegative numbers.");
System.out.println("Mark the end of the list with a negative number.");
Scanner keyboard = new Scanner(System.in);
int next;
int index = 0;
next = keyboard.nextInt();
while ((next >= 0) && (index < a.length)) {
numberUsed++;
a[index] = next;
index++;
// Print out each value of next
System.out.println(next);
next = keyboard.nextInt();
// System.out.println("Number of indexes used" + numberUsed);
}
keyboard.close(); // close the keyboard so it can't continue to be used.
return index;
}
public Integer[] sort(final Integer[] arrays) {
Arrays.sort(arrays, new Comparator<Integer>() {
@Override
public int compare(Integer int1, Integer int2) {
if (null != int1 && null != int2) {
if (int1 < int2) {
return -1;
} else if (int1 > int2) {
return 1;
} else {
return 0;
}
}
return 0;
}
});
return arrays;
}
public void count(Integer[] a) {
// for each element in array go through if conditons.
System.out.println("N " + "Count");
for (int i = 0; i < a.length - 1; i++) {
if (i == 0 && a[i] != null) {
System.out.println(a[i] + " " + count);
}
if (i > 0 && (a[i] != null && a[i - 1] != null)) {
if (a[i] == a[i - 1]) {
count++;
}
if (a[i] != a[i - 1]) {
count = 1;
}
if (a[i] != 0) {
System.out.println(a[i] + " " + count);
}
} else {
count = 1;
}
}
}
public static void main(String[] args) {
Prac_ExeOne score = new Prac_ExeOne();
score.fillArray(Array);
score.sort(Array);
score.count(Array);
}
}
结果：
输入最多50个非负数。
使用负数标记列表末尾。
3
3
2
2
-1
N计数
2 1
2 2
3 1
3 2
由于您的排序方法，您的代码无效。 （抱歉我的英语不好）