1. 汽水瓶问题

这玩意儿问题通常是一个轻巧松的数学问题，涉及计算Neng兑换的汽水瓶数量。下面是补全后的代码：

java public class DrinkBottle { public static void main { Scanner in = new Scanner; int n; while ) { n = in.nextInt; if { // 特殊情况处理 break; } int count = 0; while { // 每两个空瓶Neng换一个汽水 count += n / 2; n = n / 2 + n % 2; // 剩余的空瓶加上新鲜得到的空瓶 } if { // 再说说一个空瓶也Neng换一个汽水 count++; } System.out.println; } in.close; } }

2. 合并排序问题

这玩意儿问题是一个经典的排序算法问题， 下面是补全后的代码：

java import java.util.Arrays;

public class MergeSort { public void mergeSort { if { return; } int mid = / 2; mergeSort; mergeSort; merge; }

public void merge {
    int temp = new int;
    int i = left, j = mid + 1, k = 0;
    while  {
        if  {
            temp = nums;
        } else {
            temp = nums;
        }
    }
    while  {
        temp = nums;
    }
    while  {
        temp = nums;
    }
    for  {
        nums = temp;
    }
}
public static void main {
    int nums = {12, 11, 13, 5, 6, 7};
    MergeSort ms = new MergeSort;
    ms.mergeSort;
    System.out.println);
}

}

3. 反转链表问题

这玩意儿问题是反转一个单链表，下面是补全后的代码：

java class ListNode { int val; ListNode next; ListNode { val = x; } }

public class ReverseList { public ListNode reverseList { if { return head; } ListNode p = head.next; head.next = null; while { ListNode q = p.next; p.next = head; head = p; p = q; } return head; }

// 检查点
public static void main {
    ListNode p1 = new ListNode;
    ListNode p2 = new ListNode;
    ListNode p3 = new ListNode;
    p1.next = p2;
    p2.next = p3;
    p3.next = null;
    ReverseList test = new ReverseList;
    ListNode res = test.reverseList;
    if  {
        System.out.println;
    } else {
        System.out.println;
    }
    // 测试用例
    ListNode p4 = new ListNode;
    ListNode p5 = new ListNode;
    ListNode p6 = new ListNode;
    p4.next = p5;
    p5.next = p6;
    p6.next = null;
    ReverseList test2 = new ReverseList;
    ListNode res2 = test2.reverseList;
    if  {
        System.out.println;
    } else {
        System.out.println;
    }
}