【leetcode - 21】合并升序链表

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：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

 

1. 考虑递归，如果遍历的时候，l1节点值小于l2，则返回的应该为此时的l1节点及l1.next和l2拼接后的结果，这就形成了递归的条件：

# 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: ListNode, l2: ListNode) -> ListNode:
        if not l1:
            return l2
        if not l2:
            return l1
        if l1.val<=l2.val:
            #将l1的next指向剩下的合并后的结果
            l1.next = self.mergeTwoLists(l1.next,l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1,l2.next)
            return l2
        

2. 遍历判断

# 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: ListNode, l2: ListNode) -> ListNode:
        #最容易想到的，两个指针，遍历后判断大小
        if not l1:
            return l2
        if not l2:
            return l1
        #初始化
        head = new = ListNode(None)
        while l1 and l2:
            l1val = l1.val 
            l2val = l2.val
            if l1val<=l2val:
                head.next = ListNode(l1val)
                l1 = l1.next 
            else:
                head.next = ListNode(l2val)
                l2 = l2.next
            head = head.next
　　　　 #重点在下面，如果跳出while说明l1,l2中存在None了，直接拼接剩下的链表就好了
        head.next = l1 if l1 else l2
        return new.next