021.Merge Two Sorted Lists

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Test case

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[][]
[1][]
[][1]
[-1,0,5][1,2,3]

Solution 1: accepted

Well…this is a 2ms solution, but only beats 2%! We have to do more!

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public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode temp = result;
        while (l1 != null || l2 != null) {
            if (l1 == null) {
                temp.next = new ListNode(l2.val);
                l2 = l2.next == null? null : l2.next;
            } else if (l2 == null) {
                temp.next = new ListNode(l1.val);
                l1 = l1.next == null? null : l1.next;
            } else if (l1.val >= l2.val) {
                temp.next = new ListNode(l2.val);
                l2 = l2.next == null? null : l2.next;
            } else if (l1.val < l2.val) {
                temp.next = new ListNode(l1.val);
                l1 = l1.next == null? null : l1.next;
            }
            temp = temp.next;
        }
        
        return result.next;
    }
}

Solution 2: accepted

We only need to concatenate what’s left to the final result. This will significantly improve the performance when one list is much longer than the other.

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public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode temp = result;
        while (l1 != null && l2 != null) {
            if (l1.val >= l2.val) {
                temp.next = new ListNode(l2.val);
                l2 = l2.next == null? null : l2.next;
            } else if (l1.val < l2.val) {
                temp.next = new ListNode(l1.val);
                l1 = l1.next == null? null : l1.next;
            }
            temp = temp.next;
        }
        if (l1 != null)
            temp.next = l1;
        else if (l2 != null)
            temp.next = l2;
        
        return result.next;
    }
}