# 反转链表
# 输入一个链表，反转链表后，输出新链表的表头。
from tool import ListNode, createLinkList, printList
class Solution:
# 返回ListNode
def ReverseList(self, pHead):
# 头插法迭代
newHead = None
while pHead:
next = pHead.next # 临时节点，暂存当前节点的下一节点
pHead.next = newHead # 原地逆置
newHead = pHead # 新头结点前移
pHead = next # 原头结点后移
return newHead
if __name__ == '__main__':
s = Solution()
ll = createLinkList([1, 2, 3])
printList(s.ReverseList(ll))
def mergeTwoLists2(self, l1, l2):
"""
迭代
时间复杂度：O(m+n),空间复杂度：O(1)
"""
newHead = ListNode(None)
cur = newHead
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
cur = cur.next
l1 = l1.next
else:
cur.next = l2
cur = cur.next
l2 = l2.next
if l1:
cur.next = l1
if l2:
cur.next = l2
return newHead.next
def mergeTwoLists(self, l1, l2):
return self.mergeTwoLists2(l1, l2)
if __name__ == '__main__':
s = Solution()
l1 = createLinkList([1,2,4])
l2 = createLinkList([1,3,4])
printList(s.mergeTwoLists(l1, l2))
return head
# 哑结点
dummy = ListNode(None)
dummy.next = head
preNodeM = dummy
for i in range(m - 1):
preNodeM = preNodeM.next
head = preNodeM.next
new_head = None
new_tail = head
for i in range(n - m + 1):
next = head.next
head.next = new_head
new_head = head
head = next
preNodeM.next = new_head # 将逆置段的前一个节点 与 逆转后的链表头部相连
new_tail.next = head # 将逆转后的链表尾部 与 逆置段的后一个节点相连
return dummy.next
def reverseBetween(self, head, m, n):
return self.reverseBetween2(head, m, n)
if __name__ == '__main__':
head = createLinkList("12345")
s = Solution()
printList(s.reverseBetween(head, 2, 2))
small = []
# 建立头结点最小堆
x = 0
for i in lists:
if i:
heapq.heappush(
small, (i.val, x, i)
) # 比较tuple成员，若第一项相同，则会比较下一项。而ListNode没有比较方法，py3就会抛出异常。添加x，使比较继续。
x += 1
# 构建新有序链表
head = ptr = ListNode(None)
while small:
val, x, idx = heapq.heappop(small)
ptr.next = idx
ptr = ptr.next
idx = idx.next
if idx:
heapq.heappush(small, (idx.val, x, idx))
return head.next
def mergeKLists(self, lists):
return self.mergeKLists3(lists)
if __name__ == '__main__':
s = Solution()
l1 = createLinkList([1, 4, 5])
l2 = createLinkList([1, 3, 4])
l3 = createLinkList([2, 6])
printList(s.mergeKLists([l1, l2, l3]))
class Solution(object):
def partition(self, head, x):
"""
:type head: ListNode
:type x: int
:rtype: ListNode
"""
lesshead = ListNode(None)
lessptr = lesshead
morehead = ListNode(None)
moreptr = morehead
while head:
if head.val < x:
lessptr.next = head
lessptr = lessptr.next
else:
moreptr.next = head
moreptr = moreptr.next
head = head.next
moreptr.next = None # 将链表尾置零，否则若最后一个是比x小的，会导致成为带环链表
lessptr.next = morehead.next
return lesshead.next
if __name__ == '__main__':
head = createLinkList([1,4,3,2,5])
s = Solution()
printList(s.partition(head, 3))
if pHead1 != None:
tail.next = pHead1
if pHead2 != None:
tail.next = pHead2
return dummy.next
def Merge2(self, pHead1, pHead2):
# 递归
if not pHead1:
return pHead2
if not pHead2:
return pHead1
dummy = None
if pHead1.val < pHead2.val:
dummy = pHead1
dummy.next = self.Merge2(pHead1.next, pHead2)
else:
dummy = pHead2
dummy.next = self.Merge2(pHead1, pHead2.next)
return dummy
def Merge(self, pHead1, pHead2):
return self.Merge2(pHead1, pHead2)
if __name__ == '__main__':
s = Solution()
pHead1 = createLinkList([1, 3, 5])
pHead2 = createLinkList([2, 4, 5])
printList(s.Merge(pHead1, pHead2))
from tool import ListNode, createLinkList, printList
class Solution(object):
# 头插法迭代
def reverseList1(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head:
return head
new_head = None
while head:
next = head.next # 临时节点，暂存当前节点的下一节点
head.next = new_head # 原地逆置
new_head = head # 新头结点前移
head = next # 原头结点后移
return new_head
def reverseList(self, head):
return self.reverseList1(head)
if __name__ == '__main__':
head = createLinkList("12345")
s = Solution()
printList(s.reverseList(head))
def getIntersectionNode3(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
pa, pb = headA, headB
# 在这里第一轮体现在pa和pb第一次到达尾部会移向另一链表的表头, 而第二轮体现在如果pa或pb相交就返回交点, 不相交最后就是None == None
while pa != pb:
pa = headB if (pa == None) else pa.next
pb = headA if (pb == None) else pb.next
return pa
def getIntersectionNode(self, headA, headB):
return self.getIntersectionNode3(headA, headB)
if __name__ == '__main__':
headA = createLinkList([4,1])
headB = createLinkList([5,0,1])
headC = createLinkList([8,4,5])
headA.next.next = headC
headB.next.next.next = headC
s = Solution()
printList(s.getIntersectionNode(headA, headB))
# t = set()
# t.add(1)
# t.add(2)
# t.add(2)
# print(t)
# print(1 in t)