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# Java Program To Merge K Sorted Linked Lists – Set 1

Given K sorted linked lists of size N each, merge them and print the sorted output.

Examples:

```Input: k = 3, n =  4
list1 = 1->3->5->7->NULL
list2 = 2->4->6->8->NULL
list3 = 0->9->10->11->NULL

Output: 0->1->2->3->4->5->6->7->8->9->10->11
Merged lists in a sorted order
where every element is greater
than the previous element.

Input: k = 3, n =  3
list1 = 1->3->7->NULL
list2 = 2->4->8->NULL
list3 = 9->10->11->NULL

Output: 1->2->3->4->7->8->9->10->11
Merged lists in a sorted order
where every element is greater
than the previous element.```

Method 1 (Simple):

Approach:
A Simple Solution is to initialize the result as the first list. Now traverse all lists starting from the second list. Insert every node of the currently traversed list into result in a sorted way.

## Java

 `// Java program to merge k sorted ` `// linked lists of size n each ` `import` `java.io.*; ` ` `  `// A Linked List node ` `class` `Node ` `{ ` `  ``int` `data; ` `  ``Node next; ` ` `  `  ``// Utility function to create  ` `  ``// a new node. ` `  ``Node(``int` `key) ` `  ``{ ` `    ``data = key; ` `    ``next = ``null``; ` `  ``} ` `} ` `class` `GFG  ` `{ ` `  ``static` `Node head; ` `  ``static` `Node temp; ` ` `  `  ``/* Function to print nodes in  ` `     ``a given linked list */` `  ``static` `void` `printList(Node node) ` `  ``{ ` `    ``while``(node != ``null``) ` `    ``{ ` `      ``System.out.print(node.data + ``" "``); ` ` `  `      ``node = node.next; ` `    ``} ` `    ``System.out.println(); ` `  ``} ` ` `  `  ``// The main function that takes an array  ` `  ``// of lists arr[0..last] and generates ` `  ``// the sorted output  ` `  ``static` `Node mergeKLists(Node arr[],  ` `                          ``int` `last) ` `  ``{ ` `    ``// Traverse form second list to last ` `    ``for` `(``int` `i = ``1``; i <= last; i++) ` `    ``{ ` `      ``while``(``true``) ` `      ``{ ` `        ``// head of both the lists, ` `        ``// 0 and ith list.   ` `        ``Node head_0 = arr[``0``]; ` `        ``Node head_i = arr[i]; ` ` `  `        ``// Break if list ended ` `        ``if` `(head_i == ``null``) ` `          ``break``; ` ` `  `        ``// Smaller than first element ` `        ``if``(head_0.data >= head_i.data) ` `        ``{ ` `          ``arr[i] = head_i.next; ` `          ``head_i.next = head_0; ` `          ``arr[``0``] = head_i; ` `        ``} ` `        ``else` `        ``{ ` `          ``// Traverse the first list ` `          ``while` `(head_0.next != ``null``) ` `          ``{ ` `            ``// Smaller than next element ` `            ``if` `(head_0.next.data >= head_i.data) ` `            ``{ ` `              ``arr[i] = head_i.next; ` `              ``head_i.next = head_0.next; ` `              ``head_0.next = head_i; ` `              ``break``; ` `            ``} ` ` `  `            ``// Go to next node ` `            ``head_0 = head_0.next; ` ` `  `            ``// If last node ` `            ``if` `(head_0.next == ``null``) ` `            ``{ ` `              ``arr[i] = head_i.next; ` `              ``head_i.next = ``null``; ` `              ``head_0.next = head_i; ` `              ``head_0.next.next = ``null``; ` `              ``break``; ` `            ``} ` `          ``} ` `        ``} ` `      ``} ` `    ``} ` `    ``return` `arr[``0``]; ` `  ``} ` ` `  `  ``// Driver code   ` `  ``public` `static` `void` `main (String[] args)  ` `  ``{ ` `    ``// Number of linked lists ` `    ``int` `k = ``3``; ` ` `  `    ``// Number of elements in each list ` `    ``int` `n = ``4``; ` ` `  `    ``// an array of pointers storing the ` `    ``// head nodes of the linked lists ` ` `  `    ``Node[] arr = ``new` `Node[k]; ` ` `  `    ``arr[``0``] = ``new` `Node(``1``); ` `    ``arr[``0``].next = ``new` `Node(``3``); ` `    ``arr[``0``].next.next = ``new` `Node(``5``); ` `    ``arr[``0``].next.next.next = ``new` `Node(``7``); ` ` `  `    ``arr[``1``] = ``new` `Node(``2``); ` `    ``arr[``1``].next = ``new` `Node(``4``); ` `    ``arr[``1``].next.next = ``new` `Node(``6``); ` `    ``arr[``1``].next.next.next = ``new` `Node(``8``); ` ` `  `    ``arr[``2``] = ``new` `Node(``0``); ` `    ``arr[``2``].next = ``new` `Node(``9``); ` `    ``arr[``2``].next.next = ``new` `Node(``10``); ` `    ``arr[``2``].next.next.next = ``new` `Node(``11``); ` ` `  `    ``// Merge all lists ` `    ``head = mergeKLists(arr, k - ``1``); ` `    ``printList(head); ` `  ``} ` `} ` `// This code is contributed by avanitrachhadiya2155 `

Output:

`0 1 2 3 4 5 6 7 8 9 10 11`

Complexity Analysis:

• Time complexity: O(nk2)
• Auxiliary Space: O(1).
As no extra space is required.

Method 2: Min Heap
A Better solution is to use Min Heap-based solution which is discussed here for arrays. The time complexity of this solution would be O(nk Log k)
Method 3: Divide and Conquer
In this post, Divide and Conquer approach is discussed. This approach doesn’t require extra space for heap and works in O(nk Log k)
It is known that merging of two linked lists can be done in O(n) time and O(n) space.

1. The idea is to pair up K lists and merge each pair in linear time using O(n) space.
2. After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.
3. Repeat the procedure until we have only one list left.

Below is the implementation of the above idea.

## Java

 `// Java program to merge k sorted  ` `// linked lists of size n each ` `public` `class` `MergeKSortedLists  ` `{ ` `    ``/* Takes two lists sorted in increasing order,  ` `       ``and merge their nodes together to make one  ` `       ``big sorted list. Below function takes O(Log n)  ` `       ``extra space for recursive calls, but it can be  ` `       ``easily modified to work with same time and  ` `       ``O(1) extra space  */` `    ``public` `static` `Node SortedMerge(Node a, Node b) ` `    ``{ ` `        ``Node result = ``null``; ` `         `  `        ``// Base cases  ` `        ``if` `(a == ``null``) ` `            ``return` `b; ` `        ``else` `if` `(b == ``null``) ` `            ``return` `a; ` ` `  `        ``// Pick either a or b, and recur  ` `        ``if` `(a.data <= b.data)  ` `        ``{ ` `            ``result = a; ` `            ``result.next = SortedMerge(a.next, b); ` `        ``} ` `        ``else`  `        ``{ ` `            ``result = b; ` `            ``result.next = SortedMerge(a, b.next); ` `        ``} ` `        ``return` `result; ` `    ``} ` ` `  `    ``// The main function that takes an array  ` `    ``// of lists arr[0..last] and generates  ` `    ``// the sorted output ` `    ``public` `static` `Node mergeKLists(Node arr[],  ` `                                   ``int` `last) ` `    ``{ ` `        ``// Repeat until only one list is left ` `        ``while` `(last != ``0``)  ` `        ``{ ` `            ``int` `i = ``0``, j = last; ` ` `  `            ``// (i, j) forms a pair ` `            ``while` `(i < j)  ` `            ``{ ` `                ``// Merge List i with List j and  ` `                ``// store merged list in List i ` `                ``arr[i] = SortedMerge(arr[i], arr[j]); ` ` `  `                ``// consider next pair ` `                ``i++; ` `                ``j--; ` ` `  `                ``// If all pairs are merged, update last ` `                ``if` `(i >= j) ` `                    ``last = j; ` `            ``} ` `        ``} ` `        ``return` `arr[``0``]; ` `    ``} ` ` `  `    ``/* Function to print nodes in a  ` `       ``given linked list */` `    ``public` `static` `void` `printList(Node node) ` `    ``{ ` `        ``while` `(node != ``null``) ` `        ``{ ` `            ``System.out.print(node.data + ``" "``); ` `            ``node = node.next; ` `        ``} ` `    ``} ` ` `  `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``// Number of linked lists ` `        ``int` `k = ``3``;  ` ` `  `        ``// Number of elements in each list ` `        ``int` `n = ``4``;  ` ` `  `        ``// An array of pointers storing the  ` `        ``// head nodes of the linked lists ` `        ``Node arr[] = ``new` `Node[k]; ` ` `  `        ``arr[``0``] = ``new` `Node(``1``); ` `        ``arr[``0``].next = ``new` `Node(``3``); ` `        ``arr[``0``].next.next = ``new` `Node(``5``); ` `        ``arr[``0``].next.next.next = ``new` `Node(``7``); ` ` `  `        ``arr[``1``] = ``new` `Node(``2``); ` `        ``arr[``1``].next = ``new` `Node(``4``); ` `        ``arr[``1``].next.next = ``new` `Node(``6``); ` `        ``arr[``1``].next.next.next = ``new` `Node(``8``); ` ` `  `        ``arr[``2``] = ``new` `Node(``0``); ` `        ``arr[``2``].next = ``new` `Node(``9``); ` `        ``arr[``2``].next.next = ``new` `Node(``10``); ` `        ``arr[``2``].next.next.next = ``new` `Node(``11``); ` ` `  `        ``// Merge all lists ` `        ``Node head = mergeKLists(arr, k - ``1``); ` `        ``printList(head); ` `    ``} ` `} ` ` `  `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node next; ` `    ``Node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `    ``} ` `} ` `// This code is contributed by Gaurav Tiwari`

Output:

`0 1 2 3 4 5 6 7 8 9 10 11`

Complexity Analysis:

Assuming N(n*k) is the total number of nodes, n is the size of each linked list, and k is the total number of linked lists.

• Time Complexity: O(N*log k) or O(n*k*log k)
As outer while loop in function mergeKLists() runs log k times and every time it processes n*k elements.
• Auxiliary Space: O(N) or O(n*k)
Because recursion is used in SortedMerge() and to merge the final 2 linked lists of size N/2, N recursive calls will be made.

Please refer complete article on Merge K sorted linked lists | Set 1 for more details!

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