leetcode之數組反轉

public class Pan {

    public static void main(String[] args) {

        int[] a = new int[5];

        a[0] = (int) (Math.random() * 100);//隨機生成1~100的浮點數，然後強轉成int

        a[1] = (int) (Math.random() * 100);

        a[2] = (int) (Math.random() * 100);

        a[3] = (int) (Math.random() * 100);

        a[4] = (int) (Math.random() * 100);

 

        System.out.println("數組中各個值是:");

        for (int i = 0; i < a.length; i++)

            System.out.print(a[i] + " ");

         

        /*思路一： 使用臨時數組*/

         

        System.out.println();

        //準備臨時數組

        int[] temp = new int[a.length];

        //把原數組的內容複製給臨時數組

        for (int i = 0; i < temp.length; i++) {

            temp[i] = a[i];

        }

        System.out.println("臨時數組中的各個值是:");

        for (int i = 0; i < temp.length; i++)

            System.out.print(temp[i] + " ");

        System.out.println();

        //反轉的做法是把臨時數組的數據，挨個 倒 放入 原數組中

        for (int i = 0; i < temp.length; i++) {

            a[i] = temp[temp.length-1-i];

        }

  

        System.out.println("反轉後的數組中各個值是:");

        for (int i = 0; i < a.length; i++)

            System.out.print(a[i] + " ");

         

        System.out.println();

         

        /*思路二： 進行首尾調換*/      

        for (int i = 0; i < a.length/2; i++) {

            int middle = a[a.length-i-1];

            a[a.length-i-1] = a[i];

            a[i] = middle;

        }       

        System.out.println("再次反轉後的數組中各個值是:");

        for (int i = 0; i < a.length; i++) {

            System.out.print(a[i] + " ");

        }

        System.out.println();

         

    }

}