链表
NC 33 合并有序链表
题目描述：将两个有序的链表合并为一个新链表，要求新的链表是通过拼接两个链表的节点来生成的，且合并后新链表依然有序。
示例1
输入--{1},{2}
返回值--{1,2}
示例2
输入--{2},{1}
返回值--{1,2}
1 # class ListNode: 2 # def __init__(self, x): 3 # self.val = x 4 # self.next = None 5 6 # 7 # 8 # @param l1 ListNode类 9 # @param l2 ListNode类 10 # @return ListNode类 11 # 12 class Solution: 13 def mergeTwoLists(self , l1 , l2 ): 14 # write code here 15 if not l1: 16 return l2 17 if not l2: 18 return l1 19 20 fake_node = ListNode(0) 21 phead = fake_node 22 phead1 = l1 23 phead2 = l2 24 25 while phead1 and phead2: 26 if phead1.val < phead2.val: 27 phead.next = phead1 28 phead1 = phead1.next 29 else: 30 phead.next = phead2 31 phead2 = phead2.next 32 phead = phead.next 33 34 while phead1: 35 phead.next = phead1 36 phead1 = phead1.next 37 phead = phead.next 38 39 while phead2: 40 phead.next = phead2 41 phead2 = phead2.next 42 phead = phead.next 43 44 return fake_node.next 45
NC51 合并k个已排序的链表
题目描述
合并 k 个已排序的链表并将其作为一个已排序的链表返回。分析并描述其复杂度。
示例1
输入--[{1,2,3},{4,5,6,7}]
返回值--{1,2,3,4,5,6,7}
1 # class ListNode: 2 # def __init__(self, x): 3 # self.val = x 4 # self.next = None 5 6 # 7 # 8 # @param lists ListNode类一维数组 9 # @return ListNode类 10 # 11 class Solution: 12 13 def mergeKLists(self , lists ): 14 # write code here 15 lists_len = len(lists) 16 if lists_len == 0: 17 return None 18 elif lists_len == 1: 19 return lists[0] 20 21 first_head = self._mergeTwoLists(lists[0], lists[1]) 22 for index in range(2, len(lists)): 23 temp_head = self._mergeTwoLists(first_head, lists[index]) 24 first_head = temp_head 25 26 return first_head 27 28 def _mergeTwoLists(self , l1 , l2 ): 29 # write code here 30 if not l1: 31 return l2 32 if not l2: 33 return l1 34 35 fake_node = ListNode(0) 36 phead = fake_node 37 phead1 = l1 38 phead2 = l2 39 40 while phead1 and phead2: 41 if phead1.val < phead2.val: 42 phead.next = phead1 43 phead1 = phead1.next 44 else: 45 phead.next = phead2 46 phead2 = phead2.next 47 phead = phead.next 48 49 while phead1: 50 phead.next = phead1 51 phead1 = phead1.next 52 phead = phead.next 53 54 while phead2: 55 phead.next = phead2 56 phead2 = phead2.next 57 phead = phead.next 58 59 return fake_node.next
NC 78 反转链表
题目描述
输入一个链表，反转链表后，输出新链表的表头。
示例1
输入 {1,2,3}
返回值 {3,2,1}
1 # -*- coding:utf-8 -*- 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 class Solution: 7 # 返回ListNode 8 def ReverseList(self, pHead): 9 # write code here 10 back = None 11 front = pHead 12 while front != None: 13 temp = front 14 front = front.next 15 temp.next = back 16 back = temp 17 18 return back
NC50 链表中的节点每k个一组翻转
题目描述
将给出的链表中的节点每\ k k 个一组翻转，返回翻转后的链表
如果链表中的节点数不是\ k k 的倍数，将最后剩下的节点保持原样
你不能更改节点中的值，只能更改节点本身。
要求空间复杂度 \ O(1) O(1)
如果链表中的节点数不是\ k k 的倍数，将最后剩下的节点保持原样
你不能更改节点中的值，只能更改节点本身。
要求空间复杂度 \ O(1) O(1)
例如：
给定的链表是1\to2\to3\to4\to51→2→3→4→5
对于 \ k = 2 k=2, 你应该返回 2\to 1\to 4\to 3\to 52→1→4→3→5
对于 \ k = 3 k=3, 你应该返回 3\to2 \to1 \to 4\to 53→2→1→4→5
1 # class ListNode: 2 # def __init__(self, x): 3 # self.val = x 4 # self.next = None 5 6 # @param head ListNode类 7 # @param k int整型 8 # @return ListNode类 9 # 10 class Solution: 11 def reverseKGroup(self , head , k ): 12 k_th_node = head 13 for i in range(k): 14 if not k_th_node: 15 return head 16 k_th_node = k_th_node.next 17 18 new_head = self._reverse(head, k_th_node) 19 head.next = self.reverseKGroup(k_th_node, k) 20 21 return new_head 22 23 def _reverse(self, head, tail): 24 back = None 25 front = head 26 while front != tail: 27 temp = front 28 front = front.next 29 temp.next = back 30 back = temp 31 32 return back
NC24 删除有序链表中重复出现的元素
题目描述
给出一个升序排序的链表，删除链表中的所有重复出现的元素，只保留原链表中只出现一次的元素。
例如：
给出的链表为1 \to 2\to 3\to 3\to 4\to 4\to51→2→3→3→4→4→5, 返回1\to 2\to51→2→5.
给出的链表为1\to1 \to 1\to 2 \to 31→1→1→2→3, 返回2\to 32→3.
例如：
给出的链表为1 \to 2\to 3\to 3\to 4\to 4\to51→2→3→3→4→4→5, 返回1\to 2\to51→2→5.
给出的链表为1\to1 \to 1\to 2 \to 31→1→1→2→3, 返回2\to 32→3.
示例1
输入:{1,2,2}
返回值:{1}
1 # class ListNode: 2 # def __init__(self, x): 3 # self.val = x 4 # self.next = None 5 6 # 7 # 8 # @param head ListNode类 9 # @return ListNode类 10 # 11 class Solution: 12 def deleteDuplicates(self , head ): 13 # write code here 14 if not head: 15 return None 16 fake_node = ListNode(0) 17 back = fake_node 18 phead = head 19 20 while phead: 21 left = phead 22 compare_value = left.val 23 right = left 24 25 while right: 26 if right.val == compare_value: 27 right = right.next 28 else: 29 break 30 31 if right == left.next: 32 back.next = left 33 back = back.next 34 phead = right 35 else: 36 phead = right 37 38 back.next = None 39 40 return fake_node.next
