Remove Nth Node From End of List

Given the head of a linked list, remove the nth node from the end of the list and return its head.

O(N) Time and O(1) Space Complexity

• Get the length of the linked list.
• if n = length then return head->next
• n = length - n
• go to n-1th node then set its next to its next->next.
• delete the space of removed node, else it will cause memory leak.
• return head

Code

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(n==0 || head == NULL) return head;
        int len = 0;
        ListNode *temp = head;
        while (temp) {
            len++;
            temp = temp->next;
        }
        if (n == len) return head->next;

        n = len - n;
        temp = head;
        while (--n) {
            temp = temp->next;
        }
        ListNode* deleteNode=temp->next;
        temp->next = temp->next->next;
        delete deleteNode;
        return head;
    }
};

O(N) Time and O(1) Space Complexity, two pointers

• move fast pointer n times.
• if fast is null return head->next
• now move slow and fast pointers together until fast->next != NULL.
• Slow->next is the node to be deleted, so store it for deletion.
• move slow->next = slow->next->next
• Delete deleted node.
• NOTE: This method have same complexity as above method.

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* fast = head,*slow = head;
        if(n==0 || head == NULL) return head;
        while(n--) fast = fast->next;
        if(!fast) return head->next;
        while(fast->next){
            fast = fast->next;
            slow = slow->next;
        }
        ListNode* deleteNode=slow->next;
        slow->next = slow->next->next;
        delete deleteNode;
        return head;
    }
};