21. Merge Two Sorted Lists
#linked-list
Last updated
#linked-list
Last updated
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
// n -> no.of nodes in l1
// m -> no.of nodes in l2
// Time complexity: O(n+m)
// Space complexity:O(1), Auxillary space complexity
if(l1 == null || l2 == null) { // If any of the list is null
return l1 == null ? l2 : l1; // Return the other list
}
ListNode dummy = new ListNode(0); // Creating a dummy node
ListNode head = dummy; // Assigning dummy node to head
while(l1 != null && l2 != null) {
if(l1.val <= l2.val) { // Condition two check sorted order
// Asssign the lowest value node b/w l1 and l2 to dummy.next
dummy.next = l1;
// Move forward with l1 node, as the current one is processed
l1 = l1.next;
} else {
// The above inference can be implied here too
dummy.next = l2;
l2 = l2.next;
}
// Move forward with dummy, so that the l1 and l2 nodes are appended
dummy = dummy.next;
}
// Check if any list is not yet processed, then append them to dummy
dummy.next = l1 != null ? l1 : l2;
// In the beginning, we assign he head as dummy pointer
// The first node of head is dummy(0), so we need to return the head.next
return head.next;
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null || l2 == null) {
return l1==null ? l2 : l1;
}
if(l1.val <= l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}