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Linked List

Reverse a link list without using another

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How to reverse a link list without using another link list?

Note: Linked list is singly linked list (there is only next pointer no previous pointer).
See Solution: Reverse a link list without using another linked list

Take two pointers previous and current, increment both pointers by one till current reaches at the end of the list.

Reverse a Linked List

Reverse a Linked List

Node* reverse(Node* node)
{
	if(node == null)
		return null;
	Node * prev = node;
	Node* curr = node->next;
	
	while(curr != null){
		Node* tmp = curr->next;
		curr->next = prev;
		prev = curr;
		curr = temp;
	}
	return prev;
}

Detect if there is a cycle in a linked list

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How can you find if a given linked list has a cycle in it?

Note: Linked list is singly linked list (there is only next pointer no previous pointer).
See Solution: Detect if there is a cycle in the linked list

This can also be solved by using two pointers as we saw in Find nth element from end of linked list and Find middle element of linked list. Take two pointers pointer1 and pointer2, both pointing to head of the linked list initially, now increment pointer1 by one node and pointer2 by two nodes, If at any point of time pointer1 and pointer2 becomes equal then there is a cycle in the linked list, else pointer2 will reach end of the list and will become null eventually.

Code in C

bool isCyclePresent(Node* node){
	Node* pointer1 = node;
	Node* pointer2 = node;
	
	while(pointer2 != null && pointer2->next!=null){
		pointer1 = pointer1->next;
		pointer2 = pointer2->next->next;
		if(pointer1 == pointer2)
			return true;
	}
	return false;
}

Find nth element from end of linked list

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How to find nth element from the end of the linked list?

Or in other words Given an integer ‘n’, return a pointer to the n-th node from the end of the list.
Note: Linked list is singly linked list (there is only next pointer no previous pointer).

See Solution: find nth element from end of linked list

Keep two pointers pointer1 and pointer2, both pointing to head of the linked list initially. Increment pointer1 by one node till it reaches (n+1)th node. Now increment both pointers by one node till pointer1 becomes null. At this point the node which pointer2 is pointing to will be the nth node from the last.

Code in C/C++

Node * getNthNodeFromLast(Node * node, int n){
	Node* pointer1 = node;
	Node* pointer2 = node;

	while(pointer1 != null && n > 0){
		pointer1 = pointer1->next;
		n--;
	}
	//List does not have n nodes, so return null
	if(n > 0)
		return null;
	//traverse till pointer1 becomes null
	while(pointer1 != null){
		pointer1 = pointer1->next;
		pointer2 = pointer2->next;
	}
	return pointer2;
}

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