def partition(self, head, x):
"""
:type head: ListNode
:type x: int
:rtype: ListNode
"""
if head is None:
return None
dummy_before = ListNode(0)
cur_before = dummy_before
dummy_after = ListNode(0)
cur_after = dummy_after
cur_head = head
while cur_head is not None:
if cur_head.val < x:
cur_before.next = cur_head
cur_before = cur_before.next
else:
cur_after.next = cur_head
cur_after = cur_after.next
cur_head = cur_head.next
cur_before.next = dummy_after.next
cur_after.next = None
return dummy_before.next
def test_it_returns_the_correct_node(self):
l = ListNode('1')
last = last_of_list(l)
self.assertEqual(last.value, '1')
l = shift_list(ListNode('0'), l)
last = last_of_list(l)
self.assertEqual(last.value, '1')
def sortList(self, head: ListNode) -> ListNode:
if (head is None) or (head.next is None):
return head
# get length of linked list
n = 0
curr = head
while curr:
n += 1
curr = curr.next
header = ListNode() # dummy header node
header.next = head
step = 1
while step < n:
curr = header.next
tail = header
while curr:
left = curr
right = split(left, step)
curr = split(right, step)
tail = merge(left, right, tail)
#step *= 2
step <<= 1
return header.next
def reverseKGroup(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
dumy = ListNode(0)
dumy.next = head
cur = dumy
while cur is not None and cur.next is not None:
sub_head = cur.next
sub_tail = cur.next
sub_len = 1
for i in range(0, k - 1):
sub_tail = sub_tail.next
if sub_tail is None:
break
sub_len += 1
if sub_len == k:
next_head = sub_tail.next
new_head = self.revser_list(sub_head, k)
sub_head.next = next_head
cur.next = new_head
cur = sub_head
else:
break
return dumy.next
def sum_list(l1: Optional[ListNode[int]],
l2: Optional[ListNode[int]]) -> Optional[ListNode[int]]:
n1, n2, carry = l1, l2, 0
head, runner = None, None
while n1 or n2 or carry:
v1 = n1.val if n1 else 0
v2 = n2.val if n2 else 0
total = v1 + v2 + carry
carry = total // 10
num = total % 10
if n1:
n1 = n1.next
if n2:
n2 = n2.next
if not head:
head = ListNode(num)
runner = head
else:
runner.next = ListNode(num)
runner = runner.next
return head
def swap_nodes_in_pairs(head: ListNode) -> ListNode:
# Create a dummy node pointing to head.
dummy = ListNode(0)
dummy.next = head
current = dummy
# Iterate through the list, starting from the dummy node
# each time exchange the order of the two nodes after current.
# Stop when there is only 0 or 1 node left.
while current.next and current.next.next:
# To avoid confusion, we can rename the two nodes to be
# exchanged as node1 and node2.
node1 = current.next
node2 = current.next.next
# Exchange node1 and node2:
# current.next pointing to node2
current.next = node2
# node1.next pointing to node2.next
node1.next = node2.next
# node2.next pointing to node 1
node2.next = node1
# Move current to node1, we will exchange the two nodes
# after node1 in the next iteration.
current = node1
return dummy.next
def pad_zero(l: ListNode[int], n: int):
head = l
for _ in range(n):
node = ListNode(0)
node.next = head
head = node
return head
def partition(ll: LinkedList[T], x: T) -> LinkedList[T]:
before_head, before_tail, after_head, after_tail = None, None, None, None
# runner
n = ll.head
while n:
if n.val < x and before_head == None:
before_head = ListNode(n.val)
before_tail = before_head
elif n.val < x and before_head != None:
before_tail.next = ListNode(n.val)
before_tail = before_tail.next
if n.val >= x and after_head == None:
after_head = ListNode(n.val)
after_tail = after_head
elif n.val >= x and after_head != None:
after_tail.next = ListNode(n.val)
after_tail = after_tail.next
n = n.next
if not before_tail:
return LinkedList(after_head)
if before_tail:
before_tail.next = after_head
return LinkedList(before_head)
def insert(self, head: ListNode, x: int) -> ListNode:
if not head: # case 0: empty list
head = ListNode(x)
head.next = head
return head
pre = head
cur = head.next
while 1:
if pre.val <= x <= cur.val: # case 1: insert in middle
break
if pre.val > cur.val: # case 2: pre is tail (max), cur is head (min)
if x > pre.val or x < cur.val:
break
pre = cur
cur = cur.next
if pre == head: # case 3: we've looped around once; all elts equal
break
pre.next = ListNode(x, cur)
return head
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
dummy = ListNode(0)
p = dummy
add_on = 0
while l1 is not None or l2 is not None:
x = 0 if l1 is None else l1.val
y = 0 if l2 is None else l2.val
next_digit = x + y + add_on
if next_digit < 10:
p.next = ListNode(next_digit)
add_on = 0
else:
p.next = ListNode(next_digit % 10)
add_on = 1
l1 = l1.next if l1 is not None else None
l2 = l2.next if l2 is not None else None
p = p.next
if add_on > 0:
p.next = ListNode(1)
return dummy.next
def pad_list(head, n):
"""Pad a linked list with zeros before head"""
for i in range(n):
new_head = ListNode(0)
new_head.next = head
head = new_head
return head
def add_lists(l1, l2):
"""
Two lists contain digits stored in reverse order
"""
p = res = None
carry = 0
while l1 is not None or l2 is not None:
data1 = 0
data2 = 0
if l1 is not None:
data1 = l1.data
l1 = l1.next
if l2 is not None:
data2 = l2.data
l2 = l2.next
data = (data1 + data2 + carry) % 10
carry = 1 if data1 + data2 + carry >= 10 else 0
if res is None:
res = create_list([data])
p = res
else:
p.next = ListNode(data)
p = p.next
if carry > 0:
p.next = ListNode(1)
return res
def partition_list(head: ListNode, x: int) -> ListNode:
""" Break Down and Concatenate
We can break down the list into two parts:
1. Nodes have value smaller than x;
2. Nodes have value larger than or equal to x.
Time Complexity - O(N) - Iterate through the list once.
Space Complexty - O(1) - For pointers.
"""
# Use two dummy nodes to store two break-down lists.
s = smaller = ListNode(-1)
l = larger = ListNode(-1)
while head:
if head.val < x:
s.next = head
s = s.next
else:
l.next = head
l = l.next
head = head.next
# Last node of larger is the end.
l.next = None
# Concatenate two lists, as smaller will be followed by larger.
s.next = larger.next
return smaller.next
def merge(l1: ListNode, l2: ListNode) -> ListNode:
""" Merge two lists one element by one. """
while l1 and l2:
l1_temp = l1.next
l2_temp = l2.next
l1.next = l2
l1 = l1_temp
l2.next = l1
l2 = l2_temp
def merge_two_sorted_lists(l1: ListNode, l2: ListNode) -> ListNode:
""" Merge two sorted lists.
"""
if not l1: return l2
if not l2: return l1
if l1.val <= l2.val:
l1.next = merge_two_sorted_lists(l1.next, l2)
return l1
else:
l2.next = merge_two_sorted_lists(l1, l2.next)
return l2
def removeNthFromEnd(self, head, n):
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
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 is None:
return l2
if l2 is None:
return l1
if l1.val <= l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
l2.next = self.mergeTwoLists(l1, l2.next)
return l2
def partial_sum(ln1: Optional[ListNode[int]],
ln2: Optional[ListNode[int]]):
# the length of ln1 and ln2 is equal now
n1, n2, carry, next_sum_node = ln1, ln2, 0, None
if n1.next and n2.next:
next_sum_node, carry = partial_sum(n1.next, n2.next)
total = carry + n1.val + n2.val
carry = total // 10
sum = total % 10
sum_node = ListNode(sum)
sum_node.next = next_sum_node
return sum_node, carry
def mergeTwoLists2(self, l1: ListNode, l2: ListNode) -> ListNode:
"""
Solution #2: recursively
Time: O(n)
Space: O(1)
"""
if not l1: return l2
if not l2: return l1
if l1.val < l2.val:
l1.next = self.mergeTwoLists2(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists2(l1, l2.next)
return l2
def removeElements(self, head: ListNode, val: int) -> ListNode:
header = ListNode() # dummy header node
header.next = head
pre = header # last node that doesn't have target value
while head:
if head.val == val:
pre.next = head.next
else:
pre = head # pre.next
head = head.next
return header.next
def add_lists2(l1, l2):
len1 = get_length(l1)
len2 = get_length(l2)
if len1 < len2:
l1 = pad_list(l1, len2 - len1)
else:
l2 = pad_list(l2, len1 - len2)
t = []
carry = add_lists2_aux(l1, l2, t)
res = create_list(t)
if carry > 0:
new_head = ListNode(1)
new_head.next = res
res = new_head
return res
def insertionSortList(self, head: ListNode) -> ListNode:
if not head:
return head
header = ListNode() # dummy header node for new linked list
curr = head # the node from original list to be inserted
pre = header
while curr:
# save so we can update curr at end of loop
next = curr.next
# find where to insert
if (not pre.next) or (pre.next.val >= curr.val):
pre = header
while pre.next and pre.next.val < curr.val:
pre = pre.next
# insert between pre and pre.next
curr.next = pre.next
pre.next = curr
# update curr using saved "next"
curr = next
return header.next
def reverse_linked_list_recursion(head: ListNode) -> ListNode:
""" Reverse with recursion
We can assume the following nodes (after head) have been reversed and we
only need to reverse the current node and current.next.
It'll be hard for human beings to understand the full process of recursion,
as our brains are not designed to push many items in stack, so we need to
focus on one stage (see reverse_linked_list_single_case.jpg for reference),
rather than try to understand the full process.
For a single stage:
Given an input node head, reverse the linked list starting with head, and
return the reversed new head.
Time Complexity - O(N) - Iterate the list once.
Space Complexity - O(N) - For recursion stack.
"""
# Recursion stop condition.
if head == None or head.next == None:
return head
# Reverse the following nodes.
# temp will be the last node or the head node for the reversed list.
temp = reverse_linked_list_recursion(head.next)
# Reverse - the next node after head shall point to head now.
head.next.next = head
# Pointing head to None to avoid dead loop.
head.next = None
# Always return the new head node.
return temp
def merge_k_sorted_lists_heapq(input_lists: list) -> ListNode:
""" Solution - Min Heap
We first insert the first element (also smallest as the lists are sorted) of each
linked list in a min heap. After this, we can take out the smallest element from
the heap and add it to the merged list. After removing the smallest element from the
heap, we can insert the next element of the same list into the heap. Repeat previous
steps to populate the merged list in sorted order.
Time Complexity - O(kn*logk) - the number of elements in the priority queue will be
less than k, so the time complexity for insertion and deletion will be O(logk), there
are at most k*n elements (every node is inserted and deleted once), so the total time
complexity will be O(kn*logk)
Space Complexity - O(k) - for the priority queue (min-heap).
"""
dummy = ListNode(-1)
current = dummy
min_heap = []
# Put the root of each list in the min heap
for root in input_lists:
if root:
heapq.heappush(min_heap, root)
# Pop the smallest element from the min heap and add it to the result sorted list.
while min_heap:
node = heapq.heappop(min_heap)
current.next = node
current = current.next
# If the element poped still have next node, then add it into the heap.
if node.next:
heapq.heappush(min_heap, node.next)
return dummy.next
def add(a, b):
result = ListNode(0)
result_head = result
carry = 0
while a or b or carry:
a_digit = a.value if a else 0
b_digit = b.value if b else 0
a = a.next_node if a else None
b = b.next_node if b else None
digit_sum = a_digit + b_digit + carry
new_digit = digit_sum % 10
carry = digit_sum / 10
result.next_node = ListNode(new_digit)
result = result.next_node
return result_head.next_node
def delete_duplicates_three_pointers(head: ListNode) -> ListNode:
""" Three pointers
The solution is similar to the two pointers one. The difference is in this
method, we are using `b` pointing to head.next rather than head. So to compare
whether two nodes have same values, we need to use: a.next.val == b.val.
When the two pointers are pointing to different values, both a and b move forward
one step, while when they pointing to same values, there are some differentce.
1. a.next pointing to b now;
2. Pointer b has to handle bound problems, where b could be None, so we can't use
b = b.next directly, instead we have to see if b is already Nonem, so we won't get
an None has no next property error.
Here `a` is called `preserve` and `b` is called `current`.
Time Complexity - O(N) - Iterate the linked list once.
Space Complexity - O(1) - Constant space for pointers.
"""
# Edge cases, empty list or single element list.
if not (head and head.next): return head
# Create dummy node pointing to head
dummy = ListNode(-1)
dummy.next = head
preserve = dummy
current = head.next
# Current could be None, such as 1 -> 1.
while current:
# If the nodes have different values, move forward one step.
if preserve.next.val != current.val:
preserve = preserve.next
current = current.next
else:
# If the nodes have same values, then move current node until preseve and current
# pointing to different values.
while current and preserve.next.val == current.val:
current = current.next
# Different from two pointers solution, here a.next pointing to b.
preserve.next = current
# Current could pointing to None after the while loop before.
current = current.next if current else None
return dummy.next
def delete_duplicates(head: ListNode) -> ListNode:
""" Two pointers
To handle cases where the first node is also a duplicated node, we can
use dummy node to handle this. Firstly, we define two pointers `a` and
`b`, where `a` pointing to the dummy node, and `b` pointing to the head
node.
If the node `a` pointing to has a different value comparing to the node
`b` is pointing to, then both `a` and `b` move forward for one step. If
the values are the same, then only move `b`, and `b` keeps moving until
the values are different.
Here we are comparing a.next.val == b.next.val. While the values that
the two pointers are pointing to are equal, we can use a while loop to
keep `b` moving until we've filtered duplicated nodes.
Here `a` is called `preserve` and `b` is called `current`.
Time Complexity - O(N) - Iterate the linked list once.
Space Complexity - O(1) - Constant space for pointers.
"""
# Edge cases, empty list or single element list.
if not (head and head.next): return head
# Create dummy node pointing to head.
dummy = ListNode(0)
dummy.next = head
preserve = dummy
current = head
while current and current.next:
# If the nodes have different values, move forward one step.
if preserve.next.val != current.next.val:
preserve = preserve.next
current = current.next
else:
# If the nodes have same values, then move current node until preseve and current
# pointing to different values. Exclude all those nodes have same values.
while current and current.next and preserve.next.val == current.next.val:
current = current.next
preserve.next = current.next
current = current.next
return dummy.next
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
head = ListNode(0)
l = head
rem = 0
prevRem = 0
while l1 is not None:
val = (l1.val + l2.val) % 10
l.next = ListNode(val + prevRem)
prevRem = int(((l1.val + l2.val) - val) / 10)
l = l.next
l1 = l1.next
l2 = l2.next
return head.next
def addTwoNumbers(l1, l2):
dummy_node = ListNode(0)
p1 = l1
p2 = l2
carry = 0
curr = dummy_node
while p1 != None or p2 != None:
x = p1.val if p1 != None else 0
y = p2.val if p2 != None else 0
sum = carry + x + y
carry = sum//10
curr.next = ListNode(sum%10)
curr = curr.next
if p1 != None: p1 = p1.next
if p2 != None: p2 = p2.next
if carry > 0:
curr.next = ListNode(carry)
return dummy_node.next
def swap_pairs(head):
"""
:type head: ListNode
:rtype: ListNode
"""
node = ListNode(-1)
node.next = head
pre = node
while pre.next and pre.next.next:
l1 = pre.next
l2 = pre.next.next
nxt = l2.next
l1.next = nxt
l2.next = l1
pre.next = l2
pre = l1
return node.next
def addTwoNumbers(self, A, B):
r = ListNode(0)
head = r
rem = 0
while A is not None or B is not None:
a = 0
if A:
a = A.val
b = 0
if B:
b = B.val
v = a + b + rem
rem = 0
if v > 9:
rem = 1
v = int(str(v)[1])
r.next = ListNode(v)
r = r.next
if A:
A = A.next
if B:
B = B.next
if rem:
r.next = ListNode(rem)
r = head.next
cnt = 0
n = 0
while r is not None:
if r.val == 0:
cnt += 1
else:
cnt = 0
n += 1
r = r.next
# print n, cnt
r = head.next
if cnt:
for i in range(n - cnt - 1):
r = r.next
r.next = None
# print l2a(head.next)
return head.next
if tmp >= 10:
carryover = tmp / 10
tmp = tmp % 10
else:
carryover = 0
lcount.val = tmp
ltmp = lcount
lcount = lcount.next
if carryover > 0:
ltmp.next = ListNode(carryover)
return lhead
l1 = ListNode(2)
l1.next = ListNode(4)
l1.next.next = ListNode(3)
#print l1
l2 = ListNode(5)
l2.next = ListNode(6)
l2.next.next = ListNode(4)
#print l2
sol = Solution()
print sol.addTwoNumbers(l1,l2)
l3 = ListNode(1)
l4 = ListNode(9)
l4.next = ListNode(9)
