Middle of the Linked List

Middle of the Linked List - Problem

Given the head of a singly linked list, return the middle node of the linked list.

If there are two middle nodes, return the second middle node.

Input & Output

Example 1 — Odd Length List

$ Input: head = [1,2,3,4,5]

› Output: [3,4,5]

💡 Note: The middle node is 3. Since there's only one middle node in odd-length lists, return node 3 and all nodes after it.

Example 2 — Even Length List

$ Input: head = [1,2,3,4,5,6]

› Output: [4,5,6]

💡 Note: There are two middle nodes (3 and 4). Return the second middle node (4) and all nodes after it.

Example 3 — Single Node

$ Input: head = [1]

› Output: [1]

💡 Note: Only one node exists, so it is the middle node.

Constraints

The number of nodes in the list is in the range [1, 100]
1 ≤ Node.val ≤ 100

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15 🍎 Apple 12

The key insight is using the fast and slow pointer technique (tortoise and hare). While the two-pass approach works, the optimal solution uses two pointers moving at different speeds to find the middle in a single pass. Time: O(n), Space: O(1)

Common Approaches

✓ Two Pass - Count Then Find

⏱️ Time: O(n) Space: O(1)

First pass counts all nodes in the linked list. Second pass traverses to the middle position (length/2) to find the middle node.

Two Pointers - Fast and Slow

⏱️ Time: O(n) Space: O(1)

Fast pointer moves 2 steps at a time, slow pointer moves 1 step. When fast reaches the end, slow is at the middle. This is the classic 'tortoise and hare' algorithm.

Two Pass - Count Then Find — Algorithm Steps

Step 1: Traverse entire list to count total nodes
Step 2: Traverse again to position length/2 to find middle

Visualization

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Step-by-Step Walkthrough

First Pass

Traverse entire list to count nodes

Calculate Middle

Middle position = count / 2

Second Pass

Traverse to middle position

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct ListNode {
    int val;
    struct ListNode* next;
};

struct ListNode* solution(struct ListNode* head) {
    if (!head) return NULL;
    
    // First pass: count nodes
    int count = 0;
    struct ListNode* current = head;
    while (current) {
        count++;
        current = current->next;
    }
    
    // Second pass: find middle
    int middlePos = count / 2;
    current = head;
    for (int i = 0; i < middlePos; i++) {
        current = current->next;
    }
    
    return current;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int arr[1000];
    int size = 0;
    char* token = strtok(line, "[],\n");
    while (token) {
        arr[size++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    // Build linked list
    struct ListNode* head = NULL;
    if (size > 0) {
        head = (struct ListNode*)malloc(sizeof(struct ListNode));
        head->val = arr[0];
        head->next = NULL;
        struct ListNode* current = head;
        for (int i = 1; i < size; i++) {
            current->next = (struct ListNode*)malloc(sizeof(struct ListNode));
            current = current->next;
            current->val = arr[i];
            current->next = NULL;
        }
    }
    
    // Find middle
    struct ListNode* result = solution(head);
    
    // Find position and print result
    if (result) {
        struct ListNode* current = head;
        int pos = 0;
        while (current != result) {
            current = current->next;
            pos++;
        }
        printf("[");
        for (int i = pos; i < size; i++) {
            if (i > pos) printf(",");
            printf("%d", arr[i]);
        }
        printf("]\n");
    } else {
        printf("[]\n");
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two passes through the list: O(n) + O(n/2) = O(n)

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables for counting and traversal

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Two Pass - Count Then Find — Algorithm Steps

Visualization

Code -

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