无需在Java中使用内置排序方法即可动态排序用户输入

Question

我试图排序一些用空格分隔的用户输入的整数.

输入:4 2 1 5 9 - 预期输出:1 2 4 5 9

用户在循环中按下回车后我无法弄清楚如何停止循环,其中i <num.当我逐个输入整数时,我的代码会起作用.任何帮助,将不胜感激

 class javasort {
    public static void main(String[] args) {
    int num, i, j, temp;
    Scanner input = new Scanner(System.in);

    // System.out.println("Enter the number of integers to sort:");
    // num = input.nextInt();

    num = 5; // <-- the user input should be dynamic

    int array[] = new int[num];

    System.out.println("Enter integers: ");

    for (i = 0; i < num; i++)

        array[i] = Integer.parseInt(input.next());
        num = i; // make array as big as input ?

    for (i = 0; i < (num - 1); i++) {
        for (j = 0; j < num - i - 1; j++) {
            if (array[j] > array[j + 1]) {
                temp = array[j];
                array[j] = array[j + 1];
                array[j + 1] = temp;
            }
        }
    }

    System.out.println("Sorted list of integers:");

    for (i = 0; i < num; i++)
        System.out.println(array[i]);
}}

Answer 1

你的代码几乎是正确的,然后你删除了你的最佳提示.用Scanner.nextInt()得像

num = input.nextInt();          // <-- get the count.
int array[] = new int[num];
System.out.println("Enter integers: ");
for (i = 0; i < num; i++) {     // <-- don't rely on indentation for flow control.
    array[i] = input.nextInt(); // <-- get a number "num" times.
}