算法思想：由于单链表是单向的，想要对当前元素进行操作，需找到前一个元素。本题利用双指针，初始pre指针指向NULL，cur指针指向head.再对局部翻转之前，先把下一个结点存到temp指针中。当进行完如下代码逻辑后，此时cur指针指向NULL,pre指针指向头结点
代码
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode pre = null, cur = head;
while (cur != null) {
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
return pre;
}
}