def get_loop_linked_list(self, n):
sll = SingleLinkedList()
head = sll.add_item_from_arr(list(range(n)))
if head is None:
return None, None
ptr = head
while ptr.next:
ptr = ptr.next
ptr.next = head
return head, ptr
## 直接用list来模拟环形链表,找到删除数字之间的下标变化规律
def LastRemaining_Solution(self, n, m):
if n < 1 or m < 1:
return -1
arr = list(range(n))
index = (m - 1) % len(arr)
while len(arr) > 1:
arr.pop(index)
index = (index + m - 1) % len(arr)
return arr[0]
# 迭代解法
def ReverseList(self, pHead):
# write code here
p_reverse_head = None
cur_node = pHead
pre_node = None
while cur_node is not None:
next_node = cur_node.next ## 需要提前保存好下一个节点的值
if next_node is None:
p_reverse_head = cur_node
cur_node.next = pre_node
pre_node = cur_node
cur_node = next_node
return p_reverse_head
## 递归解法
def ReverseList(self, pHead):
if not pHead or not pHead.next:
return pHead
else:
pReversedHead = self.ReverseList(pHead.next)
pHead.next.next = pHead ## 把None换掉
pHead.next = None ## 最后要接一个None
return pReversedHead
sll = SingleLinkedList()
sll.add_item_from_arr([1, 2, 3, 4])
sol = Solution()
res = sol.ReverseList(sll.head)
pass
for i in range(k - 1): ## 在前指针移动的时候就对非正常值进行判断
if pAHead.next != None:
pAHead = pAHead.next
else:
return None
pBehind = head
while pAHead.next != None:
pAHead = pAHead.next
pBehind = pBehind.next
return pBehind
def FindMidNode(self, head): ## 这双指针的思想非常棒
if not head:
return None
pahead = head
pbehind = head
while pahead.next:
pahead = pahead.next.next
pbehind = pbehind.next
return pbehind
sll = SingleLinkedList()
sll.add_item_from_arr([1, 2, 3, 4, 5])
sll.add_item_from_arr([1])
sol = Solution()
res = sol.FindMidNode(sll.head)
pass
return count
## 解法三：调用辅助栈(利用栈后入先出的特性)
if not pHead1 or not pHead2:
return None
stack1, stack2 = [], []
while pHead1:
stack1.append(pHead1)
pHead1 = pHead1.next
while pHead2:
stack2.append(pHead2)
pHead2 = pHead2.next
res_temp1 = None
while stack1 and stack2:
temp1 = stack1.pop()
temp2 = stack2.pop()
if temp1.val != temp2.val:
return res_temp1
res_temp1 = temp1
return res_temp1
sll = SingleLinkedList()
pHead1 = sll.add_item_from_arr([1, 2, 3, 4, 5])
pHead2 = sll.add_item_from_arr([1, 2, 3, 4, 5])
sol = Solution()
res = sol.FindFirstCommonNode(pHead1, pHead2)
pass
pointer.next = pHead2
pHead2 = pHead2.next
pointer = pointer.next
if pHead1:
pointer.next = pHead1
if pHead2:
pointer.next = pHead2
return dummy.next
## 递归代码
def Merge(self, pHead1, pHead2):
if not pHead1:
return pHead2
if not pHead2:
return pHead1
if pHead1.val < pHead2.val:
pHead1.next = self.Merge(pHead1.next, pHead2)
return pHead1
else:
pHead2.next = self.Merge(pHead1, pHead2.next)
return pHead2
sll = SingleLinkedList()
pHead1 = sll.add_item_from_arr([2, 3, 6])
pHead2 = sll.add_item_from_arr([9999])
sol = Solution()
res = sol.Merge(pHead1, pHead2)
pass
