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

## A simple example of the two-pointer technique

### 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

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

#### Example 2

```
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

```
#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);
}
```

```
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

```
#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);
}
```

```
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.