leetcode 每日一题 21. 合并两个有序链表

Posted 2021-02-20 nil_f

迭代法

思路:

当L1和L2都不是空链表时，判断L1和L2哪一个链表的头节点的值更小，将较小值的节点添加到结果里。当一个节点添加到结果里后，将对应链表中的节点向后移一位。

代码：

 

# 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:
        newList = ListNode(0)
        first = newList
        while l1 and l2:
            if l1.val < l2.val:
                first.next = l1
                l1 = l1.next
            else:
                first.next = l2
                l2 = l2.next
            first = first.next
        else:
            first.next = l1 if l1 else l2
        return newList.next

 

递归法

思路：

考虑到从头开始每个节点后面都是有序的，所以可以采用递归的方法对后面的节点进行有序拼接。

代码：

# 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 not l1:
            return l2
        elif not l2:
            return l1
        elif l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next,l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1,l2.next)
            return l2