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Interview Question Merge Two Sorted Lists

Last update: 1/7/2017 7:32:00 AM


The problem is to merge two sorted list into a new list.

Problem #21 - Merge Two Sorted Lists

Description: Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Difficulty: Easy

Solution in Java

package com.fatihkabakci.Easy.MergeTwoSortedLists;

 * @author fkabakci
 * Problem Description: Merge two sorted linked lists and return it as a new list. 
 * The new list should be made by splicing together the nodes of the first two lists.
 * Solution:
 * 1. Start with the first elements. See that which one is lower than the another one, 
 * 	  then put that one (minimum one) into the list.
 * 2. Pass to the next one in the list which had a minimum one previous.
public class MergeTwoSortedLists {	
	public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
		ListNode head = new ListNode(0);
        ListNode l3 = head;
        while(l1 != null || l2 != null) {
        	int val1 = (l1 != null) ? l1.val : Integer.MAX_VALUE;
        	int val2 = (l2 != null) ? l2.val : Integer.MAX_VALUE;
        	l3.next = new ListNode(Math.min(val1, val2));
    		l3 = l3.next;
        	if(val1 < val2) 
        		l1 = l1.next;
        		l2 = l2.next;
        return head.next;
	public static void main(String[] args) {
		MergeTwoSortedLists r = new MergeTwoSortedLists();
		ListNode head = Data.init(new int[]{1, 2, 3});
		ListNode tail = Data.init(new int[]{4, 5, 6});
		ListNode entire = r.mergeTwoLists(head, tail);
		while(entire != null) {
			System.out.print(entire.val + " ->");
			entire = entire.next;

ListNode and the helper class Data which is used to initialize linked list are below.

public class ListNode {
	int val;
	ListNode next;
	ListNode(int x) {
		val = x;

package com.fatihkabakci.Easy.RemoveNthNodeFromEndofList;

public class Data {
	public static ListNode init(int[] data) {
		ListNode head = new ListNode(data[0]);
		ListNode t = head;
		for(int i = 1; i < data.length; i++) {
			t.next = new ListNode(data[i]);
			t = t.next;
		return head;

6/1/2017 8:37:00 PM


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