Leecode 206. 反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

    public ListNode reverseList(ListNode head) {
        ListNode pre = null;    //头结点变尾节点，指向null
        while(head!=null){
            ListNode next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }

    public ListNode reverseList2(ListNode head) {   //递归
        return reverseListInt(head, null);
    }
    private ListNode reverseListInt(ListNode head, ListNode pre) {
        if (head == null)
            return pre;
        ListNode next = head.next;
        head.next = pre;
        return reverseListInt(next, head);  //尾递归，操作已经完成，最后返回最后结果罢了
    }
}

 栈实现

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode resListNode = new ListNode(0) ;
        ListNode pListNode = head;
        ListNode res = resListNode;
        Stack<Integer> st =  new Stack<Integer>();
        while(pListNode!=null) {
            st.push(pListNode.val);
            pListNode = pListNode.next;
        }
        while(!st.empty()) {
            res.next = new ListNode(st.peek());
            st.pop();
            res = res.next;
        }
        return resListNode.next;
    }
}