Well, since the head
pointer may also be modified, we create a new_head
that points to it to facilitate the reverse process.
For the example list 1 -> 2 -> 3 -> 4 -> 5
in the problem statement, it will become 0 -> 1 -> 2 -> 3 -> 4 -> 5
(we init new_head -> val
to be 0
). Then we set a pointer pre
to new_head
and another cur
to head
. Then we keep inserting cur -> next
after pre
until cur becomes the last node. The code is follows.
1 class Solution { 2 public: 3 ListNode* reverseList(ListNode* head) { 4 ListNode* new_head = new ListNode(0); 5 new_head -> next = head; 6 ListNode* pre = new_head; 7 ListNode* cur = head; 8 while (cur && cur -> next) { 9 ListNode* temp = pre -> next;10 pre -> next = cur -> next;11 cur -> next = cur -> next -> next;12 pre -> next -> next = temp;13 }14 return new_head -> next;15 }16 };
provides a more concise solution without using the new_head
. The idea is to reverse one node at a time for the beginning of the list. The rewritten code is as follows.
1 class Solution { 2 public: 3 ListNode* reverseList(ListNode* head) { 4 ListNode* pre = NULL; 5 while (head) { 6 ListNode* next = head -> next; 7 head -> next = pre; 8 pre = head; 9 head = next;10 }11 return pre;12 }13 };
Well, both of the above solutions are iterative. The hint has also suggested us to use recursion. In fact, the above link has a nice recursive solution, whose rewritten code is as follows.
1 class Solution { 2 public: 3 ListNode* reverseList(ListNode* head) { 4 if (!head || !(head -> next)) return head; 5 ListNode* node = reverseList(head -> next); 6 head -> next -> next = head; 7 head -> next = NULL; 8 return node; 9 }10 };
The basic idea of this recursive solution is to reverse all the following nodes after head
. Then we need to set head
to be the final node in the reversed list. We simply set its next node in the original list (head -> next
) to point to it and sets its next
to be NULL
.