Pseudo Code
// head is the head node of a linked list
public void function(Node head):
Node slow = head;
Node fast = head;
while (fast != null && fast.next != null) { // IMPORTANT!!
slow = slow.next;
fast = fast.next.next;
}
}
- With 2 different pointers – slow and fast pointer, we can iterate through the whole list
- Note that we’re moving slow and fast pointer together, but in different speed
- The reason why we check ‘fast.next’ as well on the if statement is because fast.next.next may return NPE if fast is at the last node.
Practice Problem
class Solution {
public ListNode middleNode(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
Will the Guru 
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