Single List Reversion

LeetCode

1. 基于头插法的迭代：

 public ListNode reverseList(ListNode head) {
         if(head == null) return null;
         ListNode res = new ListNode(0);
         res.next = head;
         ListNode cur = head;
         while(cur.next != null){
             ListNode tmp = cur.next;
             cur.next = tmp.next;
             tmp.next = res.next;
             res.next = tmp;
         }
         return res.next;
 }

2. 就地逆置的简洁迭代（elegant）

 public ListNode reverse(ListNode head){
         ListNode prev = null;
         while(head != null){
             ListNode next = head.next;
             head.next = prev;
             prev = head;
             head = next;
         }
         return prev;
     }

3. 递归

 ListNode* reverseList(ListNode* head) {
         if (!head || !(head -> next)) return head;
         ListNode* node = reverseList(head -> next);
         head -> next -> next = head;
         head -> next = NULL;
         return node;
 }