面试题24：反转链表

题目：定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。
思路：定义三个参数，pre、node和next，先把node的next设置为pre，然后再往后依次操作。当node的next为null时，返回node。
解决方案：

public class Question24 {
    static class ListNode{
        int value;
        ListNode next;
        public ListNode(int value){
            this.value = value;
        }
    }
    public static ListNode reverseList(ListNode head){
        ListNode reverseHead = null;
        ListNode node = head;
        ListNode pre = null;
        while (node != null){
            ListNode next = node.next;
            if (next == null){
                reverseHead = node;
            }
            node.next = pre;
            pre = node;
            node = next;
        }
        return reverseHead;
    }

    public static void main(String[] args) {
        ListNode pHead = new ListNode(1);
        ListNode pAhead = new ListNode(3);
        ListNode pBhead = new ListNode(5);
        ListNode pChead = new ListNode(7);
        pHead.next = pAhead;
        pAhead.next = pBhead;
        pBhead.next = pChead;
        System.out.println(reverseList(pHead).value);
    }
}