Description
https://leetcode.com/problems/middle-of-the-linked-list/
Given a non-empty, singly linked list with head node head, return a middle node of linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: [1,2,3,4,5] Output: Node 3 from this list (Serialization: [3,4,5]) The returned node has value 3. (The judge's serialization of this node is [3,4,5]). Note that we returned a ListNode object ans, such that: ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6] Output: Node 4 from this list (Serialization: [4,5,6]) Since the list has two middle nodes with values 3 and 4, we return the second one.
Note:
- The number of nodes in the given list will be between
1and100.
Explanation
fast pointer to find the middle number. slow pointer to get to the middle node.
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def middleNode(self, head: ListNode) -> ListNode:
if head == None:
return head
count = 0
fast = head
while fast != None:
fast = fast.next
count += 1
mid = count // 2 + 1
slow = head
for i in range(1, mid):
slow = slow.next
return slow- Time complexity: O(n).
- Space complexity: O(1).
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode middleNode(ListNode head) {
ListNode end = head;
ListNode mid = head;
while(end!=null && end.next!=null){
end = end.next.next;
mid = mid.next;
}
return mid;
}
public ListNode middleNode1(ListNode head) {
ListNode midNode = head;
int len = getLength(head);
int mid = len/2;
while(mid > 0){
midNode = head.next;
head = head.next;
mid–;
}
return midNode;
}
public int getLength(ListNode head){
int count =0;
while(head != null){
head = head.next;
count++;
}
return count;
}
}
This looks great! To further explain this solution, I write a post with graphical explanation. Hopefully it will helps someone. It’s available at https://medium.com/@edward.zhou/leet-code-876-middle-of-the-linked-list-explained-python3-solution-a9417b418bc7.