19. Remove Nth Node From End of List

C++/19._Remove_Nth_Node_From_End_of_List

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+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode() : val(0), next(nullptr) {}
+ *     ListNode(int x) : val(x), next(nullptr) {}
+ *     ListNode(int x, ListNode *next) : val(x), next(next) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* removeNthFromEnd(ListNode* head, int n) {
+    //Optimised Method
+    //Time Complexity:- O(n) [One Pass]
+    //Space Complexity:-O(1)
+
+    //Let's keep a fast pointer and will traverse it over the list
+    //till it reaches one node before the node to be deleted.     
+        ListNode *fast=head;
+        for(int i=0;i<n;i++)
+            fast=fast->next;
+    //If fast would have reached NULL this means the node to be deleted 
+    //is equal to the number of nodes in the linked list.   
+        if(fast==NULL)
+        {
+            ListNode *node=head;
+            head=head->next;
+            delete node;
+            return head;
+        }
+    //And if not the above case, then let's keep a slow pointer which is 
+    //assigned head. Now, move both slow and fast pointer simultaneously
+    //till the fast pointer reaches the last node. Here,you will observe 
+    //that slow pointer is standing one node before the node to be 
+    //deleted.       
+        ListNode *slow=head;
+        while(fast->next!=NULL)
+        {
+            slow=slow->next;
+            fast=fast->next;
+        }
+    //Simply delete the nth node from the end of the list.         
+        ListNode *node=slow->next;
+        slow->next=node->next;
+        delete node;
+
+    return head;
+    }
+};