Python ListNode.printList方法代码示例

#Python ListNode.printList方法代码示例| 来源: 网络整理| 查看: 265

本文整理汇总了Python中ListNode.ListNode.printList方法的典型用法代码示例。如果您正苦于以下问题：Python ListNode.printList方法的具体用法？Python ListNode.printList怎么用？Python ListNode.printList使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类ListNode.ListNode的用法示例。

在下文中一共展示了ListNode.printList方法的15个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: removeDuplicatesfromUnsortedList # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Have you met this question in a real interview? Yes # Example # Given 1->3->2->1->4. # Return 1->3->2->4 # Challenge # (hard) How would you solve this problem if a temporary buffer is not allowed? # In this case, you don't need to keep the order of nodes. from ListNode import ListNode def removeDuplicatesfromUnsortedList(head): if head == None: return None dummy = ListNode(0) dummy.next = head head = dummy s = set() while head.next != None: if head.next.val in s: head.next = head.next.next else: s.add(head.next.val) head = head.next return dummy.next head = ListNode.arrayToList([1, 3, 2, 1, 4]) ListNode.printList(removeDuplicatesfromUnsortedList(head)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:removeDuplicatesfromUnsortedList.py 示例2: ListNode # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] tail = dummy carry = 0 while l1 is not None and l2 is not None: s = l1.val + l2.val + carry tail.next = ListNode(s % 10) carry = s / 10 l1 = l1.next l2 = l2.next tail = tail.next while l1 is not None: s = l1.val + carry tail.next = ListNode(s % 10) carry = s / 10 l1 = l1.next tail = tail.next while l2 is not None: s = l2.next + carry tail.next = ListNode(s % 10) carry = s / 10 l2 = l2.next tail = tail.next if carry != 0: tail.next = ListNode(carry) return dummy.next l1 = ListNode.arrayToList([6, 1, 7]) l2 = ListNode.arrayToList([2, 9, 5]) ListNode.printList(addTwoNumbersII(l1, l2)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:addTwoNumbersII.py 示例3: deleteNodeintheMiddleofSinglyLinkedList # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Implement an algorithm to delete a node in the middle of a singly linked list, # given only access to that node. # Have you met this question in a real interview? Yes # Example # Given 1->2->3->4, and node 3. return 1->2->4 from ListNode import ListNode def deleteNodeintheMiddleofSinglyLinkedList(node): if node is None: return # node is not last node if node.next is not None: node.val = node.next.val node.next = node.next.next # node is the last node else: node = None head = ListNode.arrayToList([1, 2, 3, 4]) node = head.next.next deleteNodeintheMiddleofSinglyLinkedList(node) ListNode.printList(head) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:27，代码来源:deleteNodeintheMiddleofSinglyLinkedList.py 示例4: mergeTwoSortedLists # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] from ListNode import ListNode def mergeTwoSortedLists(l1, l2): if l1 == None and l2 == None: return None dummy = ListNode(0) tail = dummy while l1 != None and l2 != None: if l1.val < l2.val: tail.next = l1 l1 = l1.next else: tail.next = l2 l2 = l2.next tail = tail.next if l1 != None: tail.next = l1 else: tail.next = l2 return dummy.next head1 = ListNode.arrayToList([1, 3, 8, 11, 15]) head2 = ListNode.arrayToList([2]) ListNode.printList(mergeTwoSortedLists(head1, head2)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:31，代码来源:mergeTwoSortedLists.py 示例5: reorderList # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Given a singly linked list L: L0 -> L1 -> ... ->Ln-1 -> Ln, # reorder it to: L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2 -> ... # You must do this in-place without altering the nodes' values. # For example, # Given 1 -> 2 -> 3 -> 4 -> null, reorder it to 1 -> 4 -> 2 -> 3 -> null. from ListNode import ListNode def reorderList(head): if head == None: return None head = ListNode.arrayToList([1, 4, 2, 3]) ListNode.printList(reorderList(head)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:19，代码来源:reorderList.py 示例6: partitionList # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # For example, # Given 1->4->3->2->5->2->null and x = 3, # return 1->2->2->4->3->5->null. from ListNode import ListNode def partitionList(head, x): # create two list:left and right leftDummy = ListNode(0) leftTail = leftDummy rightDummy = ListNode(0) rightTail = rightDummy while head != None: if head.val < x: leftTail.next = head leftTail = leftTail.next else: rightTail.next = head rightTail = rightTail.next head = head.next # combine two list rightTail.next = None leftTail.next = rightDummy.next return leftDummy.next ListNode.printList(partitionList(ListNode.arrayToList([1, 4, 3, 2, 5, 2]), 3)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:partitionList.py 示例7: removeDupfromSortedListII # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Example # Given 1->2->3->3->4->4->5, return 1->2->5. # Given 1->1->1->2->3, return 2->3. from ListNode import ListNode def removeDupfromSortedListII(head): if head == None: return None dummy = ListNode(0) dummy.next = head head = dummy while head.next != None and head.next.next != None: if head.next.val == head.next.next.val: val = head.next.val while head.next != None and head.next.val == val: head.next = head.next.next else: head = head.next return dummy.next head1 = ListNode.arrayToList([1, 2, 3, 3, 4, 4, 5]) head2 = ListNode.arrayToList([1, 1, 1, 2, 3]) ListNode.printList(removeDupfromSortedListII(head2)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:removeDuplicatesfromSortedListII.py 示例8: merge # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] slow = slow.next fast = fast.next.next return slow # helper: merge two sorted linked list def merge(head1, head2): if head1 == None and head2 == None: return None dummy = ListNode(0) lastNode = dummy while head1 != None and head2 != None: if head1.val < head2.val: lastNode.next = head1 head1 = head1.next else: lastNode.next = head2 head2 = head2.next lastNode = lastNode.next if head1 != None: lastNode.next = head1 else: lastNode.next = head2 return dummy.next head = ListNode.arrayToList([1, 3, 2, 4, 5, 7, 6]) ListNode.printList(sortList(head)) ListNode.printList(sortListI(head)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:sortList.py 示例9: removeLinkedListElements # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Remove all elements from a linked list of integers that have value val. # Have you met this question in a real interview? Yes # Example # Given 1->2->3->3->4->5->3, val = 3, you should return the list as 1->2->4->5 from ListNode import ListNode def removeLinkedListElements(head, val): if head == None or val == None: return head dummy = ListNode(0) dummy.next = head head = dummy while head.next != None: if head.next.val == val: head.next = head.next.next else: head = head.next return dummy.next head = ListNode.arrayToList([1, 2, 3, 3, 4, 5, 3]) ListNode.printList(removeLinkedListElements(head, 3)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:25，代码来源:removeLinkedListElements.py 示例10: swapNodesinPairs # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Example # Given 1->2->3->4, you should return the list as 2->1->4->3. # Challenge # Your algorithm should use only constant space. You may not modify the values # in the list, only nodes itself can be changed. from ListNode import ListNode def swapNodesinPairs(head): if head == None: return None dummy = ListNode(0) dummy.next = head head = dummy while head != None and head.next != None: n1 = head.next n2 = head.next.next # head -> n1 -> n2 ... => head -> n2 -> n1 ... head.next = n2 n1.next = n2.next n2.next = n1 # move to next pair head = n1 return dummy.next head = ListNode.arrayToList([1, 2, 3, 4]) ListNode.printList(swapNodesinPairs(head)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:swapNodesinPairs.py 示例11: mergeKHelper # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] mid = start + (end - start) / 2 left = mergeKHelper(lists, start, mid) right = mergeKHelper(lists, mid + 1, end) return mergeTwoSortedList(left, right) def mergeTwoSortedList(head1, head2): if head1 == None and head2 == None: return None dummy = ListNode(0) tail = dummy while head1 != None and head2 != None: if head1.val < head2.val: tail.next = head1 head1 = head1.next else: tail.next = head2 head2 = head2.next tail = tail.next if head1 != None: tail.next = head1 else: tail.next = head2 return dummy.next head1 = ListNode.arrayToList([2, 4]) head2 = ListNode.arrayToList([-1]) ListNode.printList(mergeKSortedLists([head1, head2])) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:32，代码来源:mergeKSortedLists.py 示例12: O # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # After removing the second node from the end, the linked list becomes # 1->2->3->5->null. # Note # The minimum number of nodes in list is n. # Challenge # O(n) time from ListNode import ListNode def removeNthNodefromEndofList(head, n): if head == None and not n: return head dummy = ListNode(0) dummy.next = head slow = dummy fast = dummy for i in range(n): fast = fast.next while fast.next != None: slow = slow.next fast = fast.next slow.next = slow.next.next return dummy.next head = ListNode.arrayToList([1, 2, 3, 4, 5]) ListNode.printList(removeNthNodefromEndofList(head, 2)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:31，代码来源:removeNthNodefromEndofList.py 示例13: range # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] head = dummy # find premmNode and mNode for i in range(1, m): head = head.next prevmNode = head mNode = head.next # reverse link from m to n nNode = mNode nextnNode = nNode.next for i in range(m, n): temp = nextnNode.next nextnNode.next = nNode nNode = nextnNode nextnNode = temp # combine prevmNode.next = nNode mNode.next = nextnNode return dummy.next head = ListNode.arrayToList([1, 2, 3, 4, 5]) ListNode.printList(reverseLinkedListII(head, 2, 4)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:28，代码来源:reverseLinkedListII.py 示例14: removeDuplicatesfromSortedList # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Given a sorted linked list, delete all duplicates such that each element # appear only once. # Have you met this question in a real interview? Yes # Example # Given 1->1->2, return 1->2. # Given 1->1->2->3->3, return 1->2->3. from ListNode import ListNode def removeDuplicatesfromSortedList(head): if head == None: return None node = head while node.next != None: if node.val == node.next.val: node.next = node.next.next else: node = node.next return head # head1 = ListNode.arrayToList([1, 1, 2]) # ListNode.printList(removeDuplicatesfromSortedList(head1)) head2 = ListNode.arrayToList([1, 1, 2, 3, 3]) ListNode.printList(removeDuplicatesfromSortedList(head2)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:29，代码来源:removeDuplicatesfromSortedList.py 示例15: reverseLinkedList # 需要导入模块: from ListNode import ListNode [as 别名] # 或者: from ListNode.ListNode import printList [as 别名] # Reverse a linked list. # Have you met this question in a real interview? Yes # Example # For linked list 1->2->3, the reversed linked list is 3->2->1 # Challenge # Reverse it in-place and in one-pass from ListNode import ListNode def reverseLinkedList(head): prev = None while head != None: temp = head.next head.next = prev prev = head head = temp return prev head = ListNode.arrayToList([1, 2, 3]) ListNode.printList(reverseLinkedList(head)) 开发者ID:cutewindy，项目名称:CodingInterview，代码行数:26，代码来源:reverseLinkedList.py

注：本文中的ListNode.ListNode.printList方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台，相关代码片段筛选自各路编程大神贡献的开源项目，源码版权归原作者所有，传播和使用请参考对应项目的License；未经允许，请勿转载。

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