5/7每日一題 (鍊結串列)

 You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

 

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

 

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

# 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]:

        dummy=ListNode() #建立一個節點,注意這個dummy是None

        tail=dummy  #尾巴從這個節點開始(另外當最後要遍歷使用)

        while list1 and list2: #當這兩個串列不為空時

            if list1.val < list2.val:

                tail.next=list1 #按照大小順序插入

                list1 = list1.next

            else:

                tail.next=list2

                list2 = list2.next

           

            tail=tail.next #不管是l1 or l2 都會指針持續移動

        if list1: #考慮l2走完l1還沒走完

            tail.next=list1

        elif list2:

            tail.next=list2

        return dummy.next #回傳起始點的下一個(因為dummy是None)