# How to solve Leetcode 876. Middle of the Linked List

### Problem statement

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.

#### Example 1

![876_lc-midlist1.jpg](https://cdn.hashnode.com/res/hashnode/image/upload/v1658767792791/0naY_SinX.jpg align="left")

```plaintext
Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.
```

#### Example 2

![876_lc-midlist2.jpg](https://cdn.hashnode.com/res/hashnode/image/upload/v1658767807055/XlxeYav-z.jpg align="left")

```plaintext
Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
```

#### Constraints

* The number of nodes in the list is in the range `[1, 100]`.
    
* `1 <= Node.val <= 100`.
    

### Solution 1: Counting the number of nodes

#### Code

```cpp
#include <iostream>
struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};
ListNode* middleNode(ListNode* head) {
    ListNode *node = head;
    int count = 0;
    while (node) {
        count++;
        node = node->next;
    }
    int i = 1;
    node = head;
    while (i <= count/2) {
        node = node->next;
        i++;
    }
    return node;        
}
void print(ListNode *head) {
    ListNode *node = head; 
    std::cout << "[";
    while (node) {
        std::cout << node->val << ",";
        node = node->next;
    }
    std::cout << "]\n";
}
int main() {
    ListNode five(5);
    ListNode four(4, &five);
    ListNode three(3, &four);    
    ListNode two(2, &three);
    ListNode one(1, &two);
    auto result = middleNode(&one);
    print(result);

    ListNode six(6);
    five.next = &six;
    result = middleNode(&one);
    print(result);
}
```

```plaintext
Output:
[3,4,5,]
[4,5,6,]
```

#### Complexity

* Runtime: `O(N + N/2)`, where `N` is the number of nodes.
    
* Extra space: `O(1)`.
    

### Solution 2: Slow and fast pointers

Use two pointers to go through the linked list.

One goes one step at a time. The other goes two steps at a time. When the faster reaches the end, the slower reaches the middle.

#### Code

```cpp
#include <iostream>
struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};
ListNode* middleNode(ListNode* head) {
    ListNode *slow = head;
    ListNode *fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;        
}
void print(ListNode *head) {
    ListNode *node = head; 
    std::cout << "[";
    while (node) {
        std::cout << node->val << ",";
        node = node->next;
    }
    std::cout << "]\n";
}
int main() {
    ListNode five(5);
    ListNode four(4, &five);
    ListNode three(3, &four);    
    ListNode two(2, &three);
    ListNode one(1, &two);
    auto result = middleNode(&one);
    print(result);

    ListNode six(6);
    five.next = &six;
    result = middleNode(&one);
    print(result);
}
```

```plaintext
Output:
[3,4,5,]
[4,5,6,]
```

#### Complexity

* Runtime: `O(N/2)`, where `N` is the number of nodes.
    
* Extra space: `O(1)`.
    

### OBS!

* The approach using slow and fast pointers looks very nice and faster. But it is not suitable to generalize this problem to any relative position (one-third, a quarter, etc.). Moreover, long expressions like `fast->next->...->next` are not recommended.
    
* Though the counting nodes approach does not seem optimized, it is more readable, scalable and maintainable.
    

### References

* [https://leetcode.com/problems/middle-of-the-linked-list/](https://leetcode.com/problems/middle-of-the-linked-list/)
    
* [https://www.leetsolve.com/876-middle-of-the-linked-list](https://www.leetsolve.com/876-middle-of-the-linked-list)
    

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