题解 | #合并两个排序的链表#

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode Merge(ListNode list1, ListNode list2) {
        if (list1 == null) return list2;
        if (list2 == null) return list1;
        ListNode dummyNode = new ListNode(-1);
        ListNode first = dummyNode;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                first.next = new ListNode(list1.val);
                list1 = list1.next;
            } else {
                first.next = new ListNode(list2.val);
                list2 = list2.next;
            }
            first = first.next;
        }

        while (list1 != null) {
            first.next = new ListNode(list1.val);
            list1 = list1.next;
            first = first.next;
        }
        while (list2 != null) {
            first.next = new ListNode(list2.val);
            list2 = list2.next;
            first = first.next;
        }
        return dummyNode.next;
    }
}

解题思想：虚拟节点+循环遍历比较+剩余节点遍历