Merge two sorted linked lists

Author: anantkaushik

Python/GeeksforGeeks/merge-two-sorted-linked-lists.py

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+"""
+Problem Link: https://practice.geeksforgeeks.org/problems/merge-two-sorted-linked-lists/1
+
+Given two sorted linked lists consisting of N and M nodes respectively. The task is to merge 
+both of the list (in-place) and return head of the merged list.
+Note: It is strongly recommended to do merging in-place using O(1) extra space.
+
+Input:
+First line of input contains number of testcases T. For each testcase, first line of 
+input contains N and M, and next two line contains N and M sorted elements in two lines for each.
+
+Output:
+For each testcase, print the merged list in sorted form.
+
+User Task:
+The task is to complete the function sortedMerge() which takes references to the heads of 
+two linked lists as the arguments and returns the head of merged linked list.
+
+Constraints:
+1 <= T <= 200
+1 <= N, M <= 103
+1 <= Node's data <= 103
+
+Example:
+Input:
+2
+4 3
+5 10 15 40 
+2 3 20
+2 2
+1 1
+2 4 
+
+Output:
+2 3 5 10 15 20 40
+1 1 2 4
+
+Explanation:
+Testcase 1: After merging the two linked lists, we have merged list as 2, 3, 5, 10, 15, 20, 40.
+"""
+#Initial Template for Python 3
+# Node Class
+class Node:
+    def __init__(self, data):   # data -> value stored in node
+        self.data = data
+        self.next = None
+# Linked List Class
+class LinkedList:
+    def __init__(self):
+        self.head = None
+    # creates a new node with given value and appends it at the end of the linked list
+    def append(self, new_value):
+        new_node = Node(new_value)
+        if self.head is None:
+            self.head = new_node
+            return
+        curr_node = self.head
+        while curr_node.next is not None:
+            curr_node = curr_node.next
+        curr_node.next = new_node
+    # prints the elements of linked list starting with head
+    def printList(self):
+        if self.head is None:
+            print(' ')
+            return
+        curr_node = self.head
+        while curr_node:
+            print(curr_node.data,end=" ")
+            curr_node=curr_node.next
+        print(' ')
+if __name__ == '__main__':
+    t=int(input())
+    for cases in range(t):
+        n,m = map(int, input().strip().split())
+        a = LinkedList() # create a new linked list 'a'.
+        b = LinkedList() # create a new linked list 'b'.
+        nodes_a = list(map(int, input().strip().split()))
+        nodes_b = list(map(int, input().strip().split()))
+        for x in nodes_a:
+            a.append(x)
+        for x in nodes_b:
+            b.append(x)
+        a.head = merge(a.head,b.head)
+        a.printList()
+''' This is a function problem.You only need to complete the function given below '''
+#User function Template for python3
+'''
+	Function to merge two sorted lists in one
+	using constant space.
+	
+	Function Arguments: head_a and head_b (head reference of both the sorted lists)
+	Return Type: head of the obtained list after merger.
+	{
+		# Node Class
+		class Node:
+		    def __init__(self, data):   # data -> value stored in node
+		        self.data = data
+		        self.next = None
+	}
+'''
+def merge(head_a,head_b):
+    #code here
+    if not head_a or not head_b:
+        return head_a or head_b
+    newHead = Node(0)
+    prev = newHead
+    while head_a and head_b:
+        if head_a.data < head_b.data:
+            prev.next = head_a
+            prev = head_a
+            head_a = head_a.next
+        else:
+            prev.next = head_b
+            prev = head_b
+            head_b = head_b.next
+    prev.next = head_a or head_b
+    return newHead.next