Python实现“合并两个有序链表”的两种方法

合并两个链表，并返回一个新的链表。新链表由前两个链表拼接而成。

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

1：将l2链表中的结点一个一个的单独插入l1链表中

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

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1 and not l2:    #l1和l2链表均不存在
            return None
        if not l1:               #l1链表不存在
            return l2
        if not l2:               #l2链表不存在
            return l1
        while l2:                #将l2链表中的结点一个一个插入l1链表中
            l2_temp_node = ListNode(l2.val)     #将l2中结点赋值为新结点，链表作为参数进行函数传递会随函数进行动态变换
            l1 = nodeInsert(l1,l2_temp_node)
            l2 = l2.next
        return l1

def nodeInsert(l1, node):
    head_node = ListNode(0)      #头指针
    head_node.next = l1
    moder_node = ListNode(0)     #指向当前结点的前一个结点
    if node.val < l1.val:      #结点小于头结点
        node.next = l1
        head_node.next = node
        return head_node.next
    moder_node = l1            #链表往后移一位
    l1 = l1.next
    while l1:                 #遍历判断链表
        if l1.val>node.val:   #判断
            node.next = l1
            moder_node.next = node
            return head_node.next
        else:          #往后遍历链表
            moder_node=l1
            l1=l1.next
    moder_node.next = node    #结点大于链表中任意结点
    return head_node.next

2：声明第三个链表作为转接

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

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1 and not l2:  # l1和l2链表均不存在
            return None
        if not l1:  # l1链表不存在
            return l2
        if not l2:  # l2链表不存在
            return l1
        l3 = ListNode(0)
        l3_head_node = l3
        while l1 is not None and l2:
            if l1.val <= l2.val:
                l3.next = l1
                l1 = l1.next
            else:
                l3.next = l2
                l2 = l2.next
            l3 = l3.next
        if l1 is not None:
            l3.next = l1
        else:
            l3.next = l2
        return l3_head_node.next

算法题来自：https://leetcode-cn.com/problems/merge-two-sorted-lists/description/