算法的解决时间和空间的问题

算法分析

运行的快慢

反转单链

 static public ListNode reverseList(ListNode head) {
   ListNode prev = null;
   ListNode curr = head;
     while (head.next !=null ){
         ListNode next = curr.next;
         curr.next = prev;

     }
     return prev;
}