dummy = Node() # X->2 X->2->7 X->1->2->7 ...
while curr is not None:
sorted_pre = dummy # 每次指向已经排好序的头部
while sorted_pre.next is not None and sorted_pre.next.val < curr.val:
# 找到当前这个节点要插入的前一个节点
sorted_pre = sorted_pre.next
# 画图比较好理解
tmp = curr.next # 记录当前剩下的节点
curr.next = sorted_pre.next # curr节点的下一个节点；连在pre的下一个
sorted_pre.next = curr # pre的下一个节点为curr节点
curr = tmp
return dummy.next
ll = LinkedList()
ll.add_first(2)
ll.add_first(3)
ll.add_first(1)
ll.add_first(5)
ll.print_list()
a = ll.head.next
b = insertSortLinkedList(a)
ll.head.next = b
ll.print_list()
# 链表排序 (归并排序 merge sort)
# 用常数空间复杂度对链表排序，时间复杂度为 O ( n log n)
def getMiddle(ll):
'''不存在dummy head 的ll'''
node1 = ll.head
node2 = ll.head
tmp = node2
while node1 is not None:
tmp = node2
node2 = node2.next
node1 = node1.next.next if node1.next is not None else None
tmp.next = None
# node2 中间节点
back = node2
front = ll.head
return front, back
ll = LinkedList()
ll.add_first(2)
ll.add_first(3)
ll.add_first(4)
ll.add_first(5)
ll.print_list()
practice_work1(ll.head.next.next)
practice_work2(ll)
# 起一个循环链表
a = Node(1)
b = Node(2)
c = Node(3)
d = Node(4)
