21. Merge Two Sorted Lists

二刷 06/2022

Version #1 Iteration

写了一个bug就是最后没有空的那段需要接到结果的尾端，因为上一个node一定是另外一个list的，所以如果不接上就断了

Time O(M + N) - M is length of list1, N is length of list2

Space O(1)

Runtime: 0 ms, faster than 100.00% of Java online submissions for Merge Two Sorted Lists.

Memory Usage: 43.3 MB, less than 28.30% of Java online submissions for Merge Two Sorted Lists.

class Solution {

    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {

        ListNode dummy = new ListNode(0);

        ListNode p = dummy;

        ListNode p1 = list1, p2 = list2;

        while (p1 != null && p2 != null) {

            if (p1.val <= p2.val) {

                p.next = p1;

                p1 = p1.next;

            } else {

                p.next = p2;

                p2 = p2.next;

            }

            p = p.next;

        }

        if (p1 != null) {

            p.next = p1;

        }

        if (p2 != null) {

            p.next = p2;

        }

        return dummy.next;

    }

}

一刷

Version #2 Recursion version
49.17 %
class Solution {
    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;
        }
    }
}

Version #1 Iteration Version
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode curr = null;
        ListNode head = null;
        if (l1.val < l2.val) {
            head = l1;
            l1 = l1.next;
        } else {
            head = l2;
            l2 = l2.next;
        }
        curr = head;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }
            curr = curr.next;
        }
        if (l1 == null) curr.next = l2;
        if (l2 == null) curr.next = l1;
        return head;
    }
}