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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
Solution
Iteration
Time complexity : O(n+m)
Space complexity : O(1)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode p = dummy;
while (l1 != null || l2 != null) {
if (l1 == null) {
p.next = l2;
break;
}
if (l2 == null) {
p.next = l1;
break;
}
if (l1.val < l2.val) {
p.next = l1;
l1 = l1.next;
} else {
p.next = l2;
l2 = l2.next;
}
p = p.next;
}
return dummy.next;
}
}
LeetCode Official - Iteration
Time complexity : O(n+m)
Space complexity : O(1)
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
// maintain an unchanging reference to node ahead of the return node.
ListNode prehead = new ListNode(-1);
ListNode prev = prehead;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
prev.next = l1;
l1 = l1.next;
} else {
prev.next = l2;
l2 = l2.next;
}
prev = prev.next;
}
// exactly one of l1 and l2 can be non-null at this point, so connect
// the non-null list to the end of the merged list.
prev.next = l1 == null ? l2 : l1;
return prehead.next;
}
}
LeetCode Official - Recursion
Time complexity : O(n+m)
Space complexity : O(n+m) - The first call to mergeTwoLists does not return until the ends of both l1 and l2 have been reached, so n+m stack frames consume O(n+m) space.