如何合并两个有序链表

【ALBB面试题】

题目：

把两个有序的链表合并成一个有序的链表，例如 1 -> 5 -> 6 和 2 -> 3 -> 7 -> 8，合并之后变成了 1 -> 2 -> 3 -> 5 -> 6 -> 7 -> 8。

解：

# 合并两个有序链表
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None


class Link:
    def __init__(self, head=None, tail=None):
        self.head = head
        self.tail = tail

    def append(self, x):
        node = Node(x)
        if self.head is None:
            self.head = node
        else:
            self.tail.next = node
        self.tail = node

link_a = Link()
link_b = Link()
for i in range(1, 8):
    if i & 1:
        link_b.append(i)
        continue
    link_a.append(i)


print("合并前：")
print("link_a: ", end=" ")
p = link_a.head
while p:
    print(p.data, end="	")
    p = p.next
print()
p = link_b.head
print("link_b: ", end=" ")
while p:
    print(p.data, end="	")
    p = p.next


def merge(link1, link2):  # 合并函数
    p1 = link1.head
    p2 = link2.head
    if not p1 or not p2:  # 为空
        return link1 if p1 else link2
    while p1 and p2:  # 不为空
        minnode = p1 if p1.data <= p2.data else p2
        if p1 is link1.head and p2 is link2.head:
            head = minnode
            tail = minnode
        else:
            tail.next = minnode
            tail = tail.next

        if p1.data < p2.data:
            p1 = p1.next
        else:
            p2 = p2.next

    if p1:
        tail.next = p1
    else:
        tail.next = p2

    newlink = Link()   # 生成新的链表
    newlink.head = head
    newlink.tail = tail
    return newlink

a = merge(link_a, link_b)
p = a.head
print()
print("合并后： ", end="")
while p:
    print(p.data, end="	")
    p = p.next