题目：

输入一个链表，反转链表后，输出新链表的表头。

解题：

思路：创建四个节点 pPrev,pNode,pNext,

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
        if(head==null)
            return null;
        ListNode newHead = null;
        ListNode pNode = head;
        ListNode pPrev = null;
        while(pNode!=null){
            ListNode pNext = pNode.next;
            if(pNext==null)
                newHead = pNode;
            pNode.next = pPrev;
            pPrev = pNode;
            pNode = pNext;
        }
        return newHead;
    }
}

标签：ListNode,val,pNode,pPrev,链表,牛客,JAVA,null
来源： https://www.cnblogs.com/yanhowever/p/12056780.html