Java 数组被无意地组织起来

Question

我正在完成一项学术作业，一切都正常，但我想知道为什么我的输出将数字按从小到大的顺序排列。

谁能告诉我在我的代码中哪里发生了这种情况？

这是我的代码:

public static void main(String[] args) {
    System.out.println("Quenten's Copy");
    Scanner input = new Scanner(System.in);

    int number[] = new int [1000000]; // initialize first array
    int count[] = new int [1000000]; // initialize second array
    int temp = 0; // intialize integer to hold for comparison
    int i; // intialize loop integer

    System.out.print("Enter seven numbers: "); // ask user for input

    // first loop - takes in user input
    for (i = 0; i < 7; i++) {
        number[i] = input.nextInt();
        }

     // second loop - checks for reoccurance of numbers
    for (i = 0; i < number.length; i++) {
        temp = number [i];
        count [temp]++;
    }

    // final loop - gives the output of the count
    for (i = 1; i < count.length; i++) {
        if (count [i] > 0) {

            // output statement - tells user each number and how many
            // times each number occurs
            System.out.printf("Number %d occurs %d times\n", i, count [i]);



        }
    }

这是输出的副本:

Enter seven numbers: 12 23 44 22 23 22 55
Number 12 occurs 1 times
Number 22 occurs 2 times
Number 23 occurs 2 times
Number 44 occurs 1 times
Number 55 occurs 1 times

正如您所看到的，它在 23 之前列出了 22，等等。我希望它按照输入的顺序打印。

Answer 1

将输入中的 int 分配给数组:

number[i] = input.nextInt();

然后你通过循环对它们进行计数:

for (i = 0; i < number.length; i++) {
    temp = number [i];
    count [temp]++;
}

如您所见，如果数字 temp是 12 ，您增加 count 的值在索引12减一。

基本上，count[i] = number of occurrence of number i

因此，当您打印 count 时数组，您会按自然顺序看到它们。

如果您想按遇到顺序打印您的项目，我建议您使用 Map存储出现的情况:

Map<Integer, Integer> count = new LinkedHashMap<>(); // LinkedHashMap remember the insert order

for (i = 0; i < number.length; i++) {
    temp = number [i];
    if (count.contains(temp)) {
        count.put(temp, count.get(temp) + 1);
    } else {
        count.put(temp, 1);
    }
}

您可以按顺序打印它们:

for (Entry<Integer, Integer> entry : count.entrySet()) {
     // entry.getKey is the number
     // entry.getValue is the occurence
}