Middle Of Linked List
Posted: 8 Dec, 2020
Given the head node of the singly linked list, return a pointer pointing to the middle of the linked list.
If there are an odd number of elements, return the middle element if there are even elements return the one which is farther from the head node.
For example, let the linked list be 1->2->3->4->null
Since the number of elements in this linked list is 4 so we have 2 middle elements, i.e. 2 and 3, but we return 3 as it is farther from the head node, i.e. 1.
Input Format :
The first line of input contains an integer ‘T’ denoting the number of test cases. The next ‘2*T’ lines represent the ‘T’ test cases. The first and only line of each test case contains integers denoting the nodes of the linked list. Each line is guaranteed to have -1 at the end to signify the end of the linked list.
Output Format :
For each test case, return a pointer pointing to the node which is at the middle of the linked list. If no midpoint exists, return a null pointer.
1.You do not need to print anything, it has already been taken care of. Just implement the given function. 2.For a linked list of size 1, the head node is the midpoint. 3.If no midpoint exists, return a null pointer.
1 <= T <= 50 0 <= N <= 4*10^4 -10^9 <= data <= 10^9 data ≠ -1 Where 'N' is the number of nodes and 'data' is the value of nodes. Time Limit: 1 sec
- If we want to know the midpoint of any linked list, it is very easy to do so if we know the number of elements in the linked list.
- We take a pointer ‘p’ and use it to traverse the linked list and until we find NULL we increment a variable ‘count’ to count the number of elements in the linked list.
- We then divide the number of elements by 2 to get the position of the middle element of the linked list.
- Finally, we traverse through the first n/2 elements of the list and return the pointer of the middle element of the linked list.
We can take the Following Approach:
- Take a pointer ‘p’ to traverse the linked list, initially pointing to head node.
- Take a variable ‘numberOfNodes’ to count the number of nodes in the linked list.
- Take a variable ‘mid’ and initialize it with ‘numberOfNodes/2’(middle of Linked List).
- Finally, take a pointer ‘ptr’ and traverse through ‘mid’ number of nodes.
- Return ‘ptr’ which is now at the middle of the linked list.
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