Single List Reversion

LeetCode

1. 基于头插法的迭代：

 1 public ListNode reverseList(ListNode head) {
 2         if(head == null) return null;
 3         ListNode res = new ListNode(0);
 4         res.next = head;
 5         ListNode cur = head;
 6         while(cur.next != null){
 7             ListNode tmp = cur.next;
 8             cur.next = tmp.next;
 9             tmp.next = res.next;
10             res.next = tmp;
11         }
12         return res.next;
13 }

2. 就地逆置的简洁迭代（elegant）

 1 public ListNode reverse(ListNode head){
 2         ListNode prev = null;
 3         while(head != null){
 4             ListNode next = head.next;
 5             head.next = prev;
 6             prev = head;
 7             head = next;
 8         }
 9         return prev;
10     }

3. 递归

1 ListNode* reverseList(ListNode* head) {
2         if (!head || !(head -> next)) return head;
3         ListNode* node = reverseList(head -> next);
4         head -> next -> next = head;
5         head -> next = NULL;
6         return node; 
7 }