21. Merge Two Sorted Lists

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

Original Solution)

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
   
    ListNode root;
    if (l1 == null) {
        return l2;
    }
    if (l2 == null) {
        return l1; 
    }
    
    if (l1.val < l2.val) {
        root = l1;
        l1 = l1.next;
    } else {
        root = l2;
        l2 = l2.next;
    }
    ListNode cur = root;
    
    while (l1 != null && l2 != null) {
        if (l1.val < l2.val) {
            cur.next = l1; 
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next; 
        }
        cur = cur.next; 
    }
    
    if (l1 != null) {
        cur.next = l1;
    } 
    
    if (l2 != null) {
        cur.next = l2; 
    }
    
    return root;
    
}

Runtime: 1 ms
Memory Usage: 38.1 MB

Some points to improve on)

• Instead of initializing the root, create a dummy and return root.next
• Use the ?: operator!

Fixed Solution)

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    if (l1 == null) {
        return l2;
    }
    if (l2 == null) {
        return l1; 
    }
    ListNode root = new ListNode(0);
    ListNode cur = root;
    
    while (l1 != null && l2 != null) {
        if (l1.val < l2.val) {
            cur.next = l1; 
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next; 
        }
        cur = cur.next; 
    }
    
    cur.next = l1 == null ? l2 : l1;
    
    return root.next;  
}

Runtime: 0 ms, faster than 100.00% of Java online submissions for Merge Two Sorted Lists.
Memory Usage: 37.8 MB, less than 99.78% of Java online submissions for Merge Two Sorted Lists.

Another solution I thought was worth noting

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
	if (l1 == null) return l2;
    if (l2 == null) return l1;
    
    if (l1.val < l2.val) {
    	l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    } else {
    	l2.next = mergeTwoLists(l1, l2.next);
        return l2;
    }
}