def sum_list_reverse(llist1, llist2):
value1, value2 = 0, 0
if llist1.head is None:
value1 = 0
if llist2.head is None:
value2 = 0
it1 = llist1.head
it2 = llist2.head
i1 = 0
i2 = 0
while it1 is not None:
value1 += it1.data * pow(10, i1)
it1 = it1.next
i1 += 1
while it2 is not None:
value2 += it2.data * pow(10, i2)
it2 = it2.next
i2 += 1
total_sum = value1 + value2
print(value1, value2)
llist = LinkedList()
# 以下这步必须先做
# llist.insert(total_sum % 10)
digits_reverse_store(llist, total_sum)
llist.print()
class Stack():
def __init__(self, size=100):
self.size = size
self.stack = LinkedList()
self.count = 0
def getCount(self):
return self.count
def isEmpty(self):
result = False
if self.count == 0:
result = True
return result
def isFull(self):
result = False
if self.count == self.size:
result = True
return result
def push(self, value):
if self.isFull():
raise IndexError("Cannot push, size of stack is full")
else:
#print(self.stack)
self.stack.insertLast(value)
self.count = self.count + 1
def pop(self):
topVal = self.top()
self.count = self.count - 1
self.stack.removeLast()
return topVal
def top(self):
if self.isEmpty():
raise IndexError("Stack is empty")
else:
topVal = self.stack.peekLast()
#topVal = self.stack[self.count - 1]
return topVal
class Queue():
def __init__(self, size=100):
self.count = 0
self.size = size
self.queue = LinkedList()
def getCount(self):
return self.count
def isEmpty(self):
result = False
if self.count == 0:
result = True
return result
def isFull(self):
result = False
if self.count == self.size:
result = True
return result
def peek(self):
if self.isEmpty():
raise IndexError("Queue is empty")
else:
frontVal = self.queue.peekFirst()
return frontVal
def enqueue(self, value):
if self.isFull():
raise IndexError("Cannot enqueue, size of queue is full")
else:
#self.queue[self.count] = value
self.queue.insertLast(value)
self.count = self.count + 1
def dequeue(self):
frontVal = self.queue.removeFirst()
self.count = self.count - 1
return frontVal
def sol_follow_up(llist1, llist2):
llist = LinkedList()
sum_follow_up(llist1.head, llist2.head, llist)
# 这个方法的瑕疵在于，这一步的操作其实应该是递归方程里边重复的一个部分，但是由于最后一步没法并入递归，所以只能在
# 外边另外写一遍了
possible_decimal = 0
if llist.head is not None:
possible_decimal = llist.head.data // 10
if possible_decimal > 0:
llist.head.data = llist.head.data % 10
llist.insert(possible_decimal)
# 注意，此时该节点变成 head node 了
llist.print()
def returnKthToLast(head, k):
fast = head
# for _ in range(1, k): #range(10) is 0~9 range(1,3) = [1,2]
# if fast:
# fast = fast.next
# else:
# return None
# let fast pointer go k step in advance
i = 1
while i <= k:
if fast:
fast = fast.next
i += 1
else:
return None
slow = head
# fast is now k nodes ahead of slow
while fast:
fast = fast.next
slow = slow.next
return slow
llist = LinkedList()
llist.initial_with_string("abcdefg") # g-f-e-d-c-b-a
LinkList.print_from_head_node(returnKthToLast(llist.head, 3))
def __init__(self, size=100):
self.count = 0
self.size = size
self.queue = LinkedList()
def __init__(self, size=100):
self.size = size
self.stack = LinkedList()
self.count = 0
llist.head.data = llist.head.data % 10
digit = (value1 + value2) + possible_decimal
llist.insert(digit) # 注意，此时该节点变成 head node 了
def sol_follow_up(llist1, llist2):
llist = LinkedList()
sum_follow_up(llist1.head, llist2.head, llist)
# 这个方法的瑕疵在于，这一步的操作其实应该是递归方程里边重复的一个部分，但是由于最后一步没法并入递归，所以只能在
# 外边另外写一遍了
possible_decimal = 0
if llist.head is not None:
possible_decimal = llist.head.data // 10
if possible_decimal > 0:
llist.head.data = llist.head.data % 10
llist.insert(possible_decimal)
# 注意，此时该节点变成 head node 了
llist.print()
l1 = LinkedList()
l1.initial_with_list((7, 1, 6))
l2 = LinkedList()
l2.initial_with_list((5, 9, 2))
l1.print()
l2.print()
sol_follow_up(l1, l2)
else:
small_tail.next = i
small_tail = i
else:
if big_head is None:
big_head = i
big_tail = i
else:
big_tail.next = i
big_tail = i
i_slow = i
i = i.next
## 这一步是最重要的。主要是把一些没用的链接马上砍掉，因为没用的链接容易导致‘环’的出现！
# i_slow.next = None
## 这一步是最重要的。主要是把一些没用的链接马上砍掉，因为没用的链接容易导致‘环’的出现！
if big_tail is not None:
big_tail.next = None
if small_tail is not None:
if big_head is not None:
small_tail.next = big_head
return small_head
else:
return big_head
from LinkList import print_from_head_node
llist = LinkedList()
llist.initial_with_string("abcdefghi") # i-h-g-f-e-d-c-b-a (del = e)
print_from_head_node(partition(llist, 'c'))