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, and 5, the middle nodes are 0, 1, 1, 2, and 2, 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

解題

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
}