Complete Day5

Day05/Add Two Numbers (Linked List).cpp

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+ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
+        ListNode* start = new ListNode();
+        ListNode* temp = start;
+        int carry = 0;
+
+        while(l1 or l2 or carry) {
+            int sum = 0;
+            if(l1) {
+                sum += l1 -> val;
+                l1 = l1 -> next;
+            }
+            if(l2) {
+                sum += l2 -> val;
+                l2 = l2 -> next;
+            }
+            sum += carry;
+            carry = sum / 10;
+            ListNode* node = new ListNode(sum % 10);
+            temp -> next = node;
+            temp = temp -> next;
+        }
+
+        return start -> next;
+    }

Day05/Delete Node in a Linked List.cpp

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+void deleteNode(ListNode* node) {
+        node -> val = node -> next -> val;
+        node -> next = node -> next -> next;
+    }

Day05/Merge Two Sorted Lists.cpp

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+ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
+        ListNode* curr1 = list1;
+        ListNode* curr2 = list2;
+
+        ListNode* ans = new ListNode();
+        ListNode* curr = ans;
+
+        while(curr1 and curr2) {
+            if(curr1 -> val < curr2 -> val) {
+                curr -> next = curr1;
+                curr1 = curr1 -> next;
+            } else {
+                curr -> next = curr2;
+                curr2 = curr2 -> next;
+            }
+            curr = curr -> next;
+        }
+
+        if(curr1) {
+            curr -> next = curr1;
+        } 
+        if(curr2) {
+            curr -> next = curr2;
+        }
+
+        return ans -> next;
+    }

Day05/README.md

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+# Day5
+
+## Reverse Linked List
+
+Link: [https://leetcode.com/problems/reverse-linked-list/]
+
+- Initialize three pointers, prev and temp as null and curr as head.
+- While the curr is not null, do: 
+  - Assign next of curr to temp.
+  - Assign prev to next of curr.
+  - Assign curr to prev
+  - Assign temp to curr
+- Return prev
+
+Time Complexity: O(n)
+
+## Middle of the Linked List
+
+Link: [https://leetcode.com/problems/middle-of-the-linked-list/]
+
+- Use two pointers slow and fast.
+```
+slow = slow -> next
+fast = fast -> next -> next
+```
+- Run the while while both fast and fast -> next are valid.
+- Return the slow pointer.
+
+Time Complexity: O(n)
+
+## Delete Node in a Linked List
+
+Link: [https://leetcode.com/problems/delete-node-in-a-linked-list/]
+
+- Change the given node's value to the value of the next node.
+- Change the next address of node to next -> next.
+```
+node -> next = node -> next -> next
+```
+
+Time Complexity: O(1)
+
+## Remove Nth Node From End of List
+
+Link: [https://leetcode.com/problems/remove-nth-node-from-end-of-list/]
+
+- Create a new dummy node start.
+- Put the head node in start -> next
+- Run a fast pointer for n times.
+- Run a slow and fast pointers while fast is not null.
+- Delete the node by marking 
+```
+slow -> next = slow -> next -> next
+```
+- Return start -> next
+
+Time Complexity: O(n)
+
+## Add Two Numbers (Linked List)
+
+Link: [https://leetcode.com/problems/add-two-numbers/]
+
+- Create a dummy node and a temp node pointing to dummy node.
+- Initialize a carry variable.
+- Run a while loop while either:
+  - l1 is not empty.
+  - l2 is not empty.
+  - Carry is not zero.
+- If l1 is not null, add it's value to sum.
+- If l2 is not null, add it's value to sum.
+- If carry is not zero, add it's value to sum.
+- Set the carry as sum / 10.
+- Create a new node with sum % 10 as the value.
+- Append the new node to the next of temp pointer.
+- Return start -> next as the result.
+
+Time Complexity: O(max(n, m))
+
+## Merge two sorted Linked Lists
+
+Link: [https://leetcode.com/problems/merge-two-sorted-lists/]
+
+## Merge Two Sorted Lists
+
+Link: [https://leetcode.com/problems/merge-two-sorted-lists/]
+
+- Create a dummy node ans and a curr pointer to ans.
+- Run a while loop while both list1 and list are not null
+- Compare the current element of the two lists and append the smaller one to the curr pointer.
+- At the end of the loop, if there are elements present in either list, append the whole list to the curr pointer.
+- Return ans -> next as the result.

Day05/Remove Nth Node From End of List.cpp

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+ListNode* removeNthFromEnd(ListNode* head, int n) {
+        ListNode* start = new ListNode();
+        start -> next = head;
+        ListNode* slow = start;
+        ListNode* fast = start;
+
+        for(int i = 0; i < n; i++) {
+            fast = fast -> next;
+        }
+
+        while(fast -> next) {
+            fast = fast -> next;
+            slow = slow -> next;
+        }
+
+        slow -> next = slow -> next -> next;
+
+        return start -> next;
+    }