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# Python Program For Merging Two Sorted Linked Lists Such That Merged List Is In Reverse Order

Given two linked lists sorted in increasing order. Merge them such a way that the result list is in decreasing order (reverse order).

Examples:

```Input:  a: 5->10->15->40
b: 2->3->20
Output: res: 40->20->15->10->5->3->2

Input:  a: NULL
b: 2->3->20
Output: res: 20->3->2```

A Simple Solution is to do following.
1) Reverse first list ‘a’
2) Reverse second list ‘b’
3) Merge two reversed lists.
Another Simple Solution is first Merge both lists, then reverse the merged list.
Both of the above solutions require two traversals of linked list.

How to solve without reverse, O(1) auxiliary space (in-place) and only one traversal of both lists?
The idea is to follow merge style process. Initialize result list as empty. Traverse both lists from beginning to end. Compare current nodes of both lists and insert smaller of two at the beginning of the result list.

```1) Initialize result list as empty: res = NULL.
2) Let 'a' and 'b' be heads first and second lists respectively.
3) While (a != NULL and b != NULL)
a) Find the smaller of two (Current 'a' and 'b')
b) Insert the smaller value node at the front of the result.
c) Move ahead in the list of the smaller nodes.
4) If 'b' becomes NULL before 'a', insert all nodes of 'a'
into the result list at the beginning.
5) If 'a' becomes NULL before 'b', insert all nodes of 'a'
into result list at the beginning. ```

Below is the implementation of above solution.

## Python3

 `# Given two sorted non-empty linked lists.  ` `# Merge them in such a way that the result  ` `# list will be in reverse order. Reversing  ` `# of linked list is not allowed. Also, ` `# extra space should be O(1)  ` ` `  `# Node of a linked list  ` `class` `Node:  ` `    ``def` `__init__(``self``, ``next` `=` `None``,  ` `                 ``data ``=` `None``):  ` `        ``self``.``next` `=` `next` `        ``self``.data ``=` `data  ` ` `  `# Given two non-empty linked lists  ` `# 'a' and 'b' ` `def` `SortedMerge(a,b): ` ` `  `    ``# If both lists are empty ` `    ``if` `(a ``=``=` `None` `and` `b ``=``=` `None``): ` `        ``return` `None` ` `  `    ``# Initialize head of the  ` `    ``# resultant list ` `    ``res ``=` `None` ` `  `    ``# Traverse both lists while both  ` `    ``# of then have nodes. ` `    ``while` `(a !``=` `None` `and` `b !``=` `None``): ` `     `  `        ``# If a's current value is smaller  ` `        ``# or equal to b's current value. ` `        ``if` `(a.key <``=` `b.key): ` `         `  `            ``# Store next of current Node  ` `            ``# in first list ` `            ``temp ``=` `a.``next` ` `  `            ``# Add 'a' at the front of  ` `            ``# resultant list ` `            ``a.``next` `=` `res ` `            ``res ``=` `a ` ` `  `            ``# Move ahead in first list ` `            ``a ``=` `temp ` `         `  `        ``# If a's value is greater. Below steps  ` `        ``# are similar to above (Only 'a' is  ` `        ``# replaced with 'b') ` `        ``else``:         ` `            ``temp ``=` `b.``next` `            ``b.``next` `=` `res ` `            ``res ``=` `b ` `            ``b ``=` `temp ` `         `  `    ``# If second list reached end, but first  ` `    ``# list has nodes. Add remaining nodes of  ` `    ``# first list at the front of result list ` `    ``while` `(a !``=` `None``):     ` `        ``temp ``=` `a.``next` `        ``a.``next` `=` `res ` `        ``res ``=` `a ` `        ``a ``=` `temp ` `     `  `    ``# If first list reached end, but second  ` `    ``# list has node. Add remaining nodes of  ` `    ``# first list at the front of result list ` `    ``while` `(b !``=` `None``): ` `     `  `        ``temp ``=` `b.``next` `        ``b.``next` `=` `res ` `        ``res ``=` `b ` `        ``b ``=` `temp ` `     `  `    ``return` `res ` ` `  `# Function to print Nodes in a given  ` `# linked list  ` `def` `printList(Node): ` ` `  `    ``while` `(Node !``=` `None``): ` `     `  `        ``print``( Node.key, end ``=` `" "``) ` `        ``Node ``=` `Node.``next` `     `  `# Utility function to create a new  ` `# node with given key  ` `def` `newNode(key): ` `    ``temp ``=` `Node() ` `    ``temp.key ``=` `key ` `    ``temp.``next` `=` `None` `    ``return` `temp ` ` `  `# Driver code ` `# Start with the empty list  ` `res ``=` `None` ` `  `# Let us create two sorted linked lists  ` `# to test the above functions. Created  ` `# lists shall be ` `#     a: 5.10.15 ` `#     b: 2.3.20  ` `a ``=` `newNode(``5``) ` `a.``next` `=` `newNode(``10``) ` `a.``next``.``next` `=` `newNode(``15``) ` ` `  `b ``=` `newNode(``2``) ` `b.``next` `=` `newNode(``3``) ` `b.``next``.``next` `=` `newNode(``20``) ` ` `  `print``(``"List A before merge: "``) ` `printList(a) ` ` `  `print``(" ` `List` `B before merge: ") ` `printList(b) ` ` `  `# Merge 2 increasing order LLs  ` `# in descresing order  ` `res ``=` `SortedMerge(a, b) ` ` `  `print``(" ` `Merged Linked ``List` `is``: ") ` `printList(res) ` `# This code is contributed by Arnab Kundu `

Output:

```List A before merge:
5 10 15
List B before merge:
2 3 20
20 15 10 5 3 2 ```

Time Complexity: O(N)

Auxiliary Space: O(1)

This solution traverses both lists only once, doesn’t require reverse and works in-place.