Given a singly linked list, delete middle of the linked list. For example, if given linked list is 1->2->3->4->5 then linked list should be modified to 1->2->4->5
If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if given linked list is 1->2->3->4->5->6 then it should be modified to 1->2->3->5->6.
If the input linked list is NULL, then it should remain NULL.
If the input linked list has 1 node, then this node should be deleted and new head should be returned.
A Simple Solution is to first count number of nodes in linked list, then delete n/2’th node using the simple deletion process.
The above solution requires two traversals of linked list. We can delete middle node using one traversal. The idea is to use two pointers, slow_ptr and fast_ptr. Both pointers start from head of list. When fast_ptr reaches end, slow_ptr reaches middle. This idea is same as the one used in method 2 of this post. The additional thing in this post is to keep track of previous of middle so that we can delete middle.
Below is the implementation.
Gven Linked List 1->2->3->4->NULL Linked List after deletion of middle 1->2->4->NULL
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