2095. Delete the Middle Node of a Linked List
Medium
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
這題的話，可以透過三個指標來解，當fast跑到最後一個節點時，slow正好是位於中間需要被刪除的點。透過將pre.Next改為slow.Next，來將slow排除於linked list外。
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func deleteMiddle(head *ListNode) *ListNode {
if head == nil || head.Next == nil { return nil }
pre, slow, fast := head, head, head
for slow.Next!=nil && fast.Next!=nil {
pre = slow
slow = slow.Next
fast = fast.Next
if fast.Next != nil {
fast = fast.Next
}
}
pre.Next = slow.Next
return head
}