21. Merge Two Sorted Lists

[ZhongKuo0228]

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.

[ZhongKuo0228]

用python寫

# 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]:
        #先做一個假的開頭出來
        curr = dummy = ListNode()
        #開始跟兩個list比
        while list1 and list2:
            if list1.val < list2.val:
                curr.next = list1
                list1 = list1.next
            else:
                curr.next = list2
                list2 = list2.next
            curr = curr.next
        #將剩下沒比完的接到合併好的Linked list中
        if list1:
            curr.next = list1
        if list2:
            curr.next = list2
        #最後回傳扣除開頭的假的node的Linked list 
        return dummy.next

[fockspaces]

# 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, list2):
        head = ListNode()
        cur = head
        while list1 and list2:
            if list1.val < list2.val:
                cur.next = list1
                list1 = list1.next
            else:
                cur.next = list2
                list2 = list2.next
            cur = cur.next
        cur.next = list1 if list1 else list2
        return head.next

[fockspaces]

assignment 可以再進一步壓縮（但沒必要）

# 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, list2):
        head = cur = ListNode()
        while list1 and list2:
            if list1.val < list2.val:
                cur.next, list1 = list1, list1.next
            else:
                cur.next, list2 = list2, list2.next
            cur = cur.next
        cur.next = list1 if list1 else list2
        return head.next