假设我们有两个排序的列表A和B。我们必须将它们合并并仅形成一个排序的列表C。列表的大小可能会有所不同。
例如,假设A = [1,2,4,7]和B = [1,3,4,5,6,8],则合并列表C将为[1,1,2,3,4, 4,5,6,7,8]
我们将使用递归解决此问题。所以该功能将如下所示工作-
假设函数列表A和B merge()
如果A为空,则返回B,如果B为空,则返回A
如果A中的值<= B中的值,则A.next = merge(A.next,B)并返回A
否则,则B.next = merge(A,B.next)并返回B
让我们看一下实现以获得更好的理解
class ListNode:
def __init__(self, data, next = None):
self.val = data
self.next = next
def make_list(elements):
head = ListNode(elements[0])
for element in elements[1:]:
ptr = head
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
return head
def print_list(head):
ptr = head
print('[', end = "")
while ptr:
print(ptr.val, end = ", ")
ptr = ptr.next
print(']')
class Solution:
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if(l1.val<=l2.val):
l1.next = self.mergeTwoLists(l1.next,l2)
return l1
else:
l2.next = self.mergeTwoLists(l1,l2.next)
return l2
head1 = make_list([1,2,4,7])
head2 = make_list([1,3,4,5,6,8])
ob1 = Solution()head3 = ob1.mergeTwoLists(head1,head2)
print_list(head3)head1 = make_list([1,2,4,7]) head2 = make_list([1,3,4,5,6,8])
输出结果
[1, 1, 2, 3, 4, 4, 5, 6, 7, 8, ]