合并两个有序链表

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]
示例 2：

输入：l1 = [], l2 = []
输出：[]
示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

两个链表的节点数目范围是 [0, 50]
-100 <= Node.val <= 100
l1 和 l2 均按 非递减顺序 排列

type ListNode struct {
    Val  int
    Next *ListNode
}

func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
    res := ListNode{}
    head := &res
    for list1 != nil || list2 != nil {
        if list1 == nil {
            head.Next = list2
            break
        }
        if list2 == nil {
            head.Next = list1
            break
        }
        if list1.Val < list2.Val {
            head.Next = list1
            list1 = list1.Next
        } else {
            head.Next = list2
            list2 = list2.Next
        }
        head = head.Next
    }
    return res.Next
}

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        res = ListNode()
        head = res
        while list1 != None or list2 != None:
            if list1 == None:
                res.next = list2
                break
            if list2 == None:
                res.next = list1
                break
            if list1.val < list2.val:
                res.next = list1
                list1 = list1.next
            else:
                res.next = list2
                list2 = list2.next
            res = res.next
        return head.next

结束！