https://leetcode.cn/problems/merge-two-sorted-lists/ (opens new window) 需要注意的点

1. 需要返回虚拟头节点.next
2. 进行迭代的是另一个指针， 需要进行不断后移
3. 判断的重要性

# 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, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        # 虚拟头节点，返回该节点.next
        prehead = ListNode(-1)
        # 进行迭代的真实指针
        prev = prehead
        p1 = l1
        p2 = l2
        while p1 and p2:
            if p1.val >= p2.val:
                prev.next = p2
                p2 = p2.next
            else:
                prev.next = p1
                p1 = p1.next
            
            # 此时的prev 的指针并不是更新后的位置，需要往后移
            # prev = prev.next
            # 可以简单理解成1. prev = prehead, 2. prev.next = p1 or p2 3.prev = prev.next
            prev = prev.next
        
        # 这个判断很重要
        # 如果p1 or p2 为空，返回p1 or p2
        # 如果进行过循环后，p1 or p2 为空, prev.next = p1 or p2
        prev.next = p1 if p1 is not None  else p2

        # 返回prehead.next
        return prehead.next