You are given the head
of a linked list. Delete the middle node, and return the head
of the modified linked list.
The middle node of a linked list of size n
is the ⌊n / 2⌋th
node from the start using 0-based indexing, where ⌊x⌋
denotes the largest integer less than or equal to x
.
- For
n
=1
,2
,3
,4
, and5
, the middle nodes are0
,1
,1
,2
, and2
, respectively.
Example 1:
Input: head = [1,3,4,7,1,2,6] Output: [1,3,4,1,2,6] Explanation: The above figure represents the given linked list. The indices of the nodes are written below. Since n = 7, node 3 with value 7 is the middle node, which is marked in red. We return the new list after removing this node.
Example 2:
Input: head = [1,2,3,4] Output: [1,2,4] Explanation: The above figure represents the given linked list. For n = 4, node 2 with value 3 is the middle node, which is marked in red.
Example 3:
Input: head = [2,1] Output: [2] Explanation: The above figure represents the given linked list. For n = 2, node 1 with value 1 is the middle node, which is marked in red. Node 0 with value 2 is the only node remaining after removing node 1.
Constraints:
- The number of nodes in the list is in the range
[1, 105]
. 1 <= Node.val <= 105
Solution: Fast / Slow pointers
Use fast / slow pointers to find the previous node of the middle one, then skip the middle one.
prev.next = prev.next.next
Time complexity: O(n)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
class Solution { public: ListNode* deleteMiddle(ListNode* head) { ListNode dummy(0, head); ListNode* prev = &dummy; ListNode* fast = head; // prev points to the previous node of the middle one. while (fast && fast->next) { prev = prev->next; fast = fast->next->next; } prev->next = prev->next->next; return dummy.next; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment