LeetCode in Python-21. Merge Two Sorted Lists 合并两个有序链表

Leetcode LeetCode

创建时间:2019-07-26 17:33

字数:215 阅读:

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题目
解法1、
解法2、
出处

题目
解法1、
解法2、
出处

题目

合并两个有序链表

解法1、

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        curr = dummy = ListNode(0)
        while l1 and l2:
            if l1.val < l2.val:
                curr.next = l1
                l1 = l1.next
            else:
                curr.next = l2
                l2 = l2.next
            curr = curr.next
        curr.next = l1 or l2
        return dummy.next
if __name__ == "__main__":
    l1 = ListNode(1)
    l1.next = ListNode(2)
    l1.next.next = ListNode(4)
    l2 = ListNode(1)
    l2.next = ListNode(3)
    l2.next.next = ListNode(4)
    r = Solution().mergeTwoLists(l1, l2)

    while r:
        print(r.val)
        r = r.next

1、两个变量指向新链表，curr用来更新链表

解法2、

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 and l2:
            if l1.val > l2.val: l1, l2 = l2, l1
            l1.next = self.mergeTwoLists(l1.next, l2)
        return l1 or l2

出处

1、https://www.bilibili.com/video/av45843264
2、对应题目下Knife丶的题解

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