PROBLEM DESCRIPTION
Given the head of a linked list, remove the nth node from the end of the list and return its head.
SOLUTION
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class Solution {
public int getListSize(ListNode node){
int size = 0;
while(node!=null){
size++;
node=node.next;
}
return size;
}
public ListNode removeNthFromEnd(ListNode head, int n) {
int size = getListSize(head);
int k = (size-n);
if(k == 0) return head.next;
ListNode temp = head;
while(k>1 && head !=null){
k--;
temp = temp.next;
}
if(temp.next != null)
temp.next = temp.next.next;
return head;
}
}
Another way to code using Dummy head (Two Pass)
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class Solution {
public int getListSize(ListNode node){
int size = 0;
while(node!=null){
size++;
node=node.next;
}
return size;
}
public ListNode removeNthFromEnd(ListNode head, int n) {
int size = getListSize(head);
int k = size-n+1; //kth node from the start
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode temp = dummyHead;
while(k>1 && temp != null){
k--;
temp = temp.next;
}
temp.next = temp.next.next;
return dummyHead.next;
}
}
Optimization (Single Pass | Two Pointers)
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class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode t1 = dummyHead;
ListNode t2 = dummyHead;
//Move t1 N steps ahead
for(int i=0; i<n; i++){
t1 = t1.next;
}
//Keep moving t1 and t2 until t1 reaches last Node in the list
//At that point, t2 will be before the Node which needs to be removed
while(t1.next != null){
t1 = t1.next;
t2 = t2.next;
}
t2.next = t2.next.next;
return dummyHead.next;
}
}