要按照链表中节点的值对链表进行排序,可以使用以下步骤:
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def mergeTwoLists(l1, l2):
if not l1:
return l2
if not l2:
return l1
if l1.val < l2.val:
l1.next = mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = mergeTwoLists(l1, l2.next)
return l2
def splitList(head):
if not head or not head.next:
return head
slow = head
fast = head.next
while fast and fast.next:
slow = slow.next
fast = fast.next.next
second_half = slow.next
slow.next = None
return head, second_half
def mergeSort(head):
if not head or not head.next:
return head
first_half, second_half = splitList(head)
first_half = mergeSort(first_half)
second_half = mergeSort(second_half)
return mergeTwoLists(first_half, second_half)
def sortList(head):
return mergeSort(head)
这样,就可以通过调用sortList(head)
函数对链表进行排序。
上一篇:按对象的属性添加到列表
下一篇:按对象分组并求和的值