#!/usr/bin/python
"""
Write a function to find the middle node of a singly-linked list.
http://nbl.cewit.stonybrook.edu/algowiki/index.php/Data-structures-TADM2E-2
"""
from linked_list import randlinkedlist
ll = randlinkedlist(size=9)
print "The linked list: {0}".format(ll)
slow, fast = ll.head, ll.head
while fast:
if fast.next is None:
break
slow, fast = slow.next, fast.next.next
print "The middle of the linked list: {0}".format(slow.value)
#!/usr/bin/python
# vim: foldlevel=0
"""
Reverse a linked list in groups of size k.
http://www.geeksforgeeks.org/reverse-a-list-in-groups-of-given-size/
"""
from linked_list import randlinkedlist
ll = randlinkedlist()
print "The linked list: {0}".format(ll)
def reverse_linked_list(head, k):
if not head:
return
node = head
prev, next = None, None
count = 0
while node and count < k:
next = node.next
node.next = prev
prev, node = node, next
count += 1
head.next = reverse_linked_list(node, k)
return prev
ll.head = reverse_linked_list(ll.head, 3)
print "The reversed linked list (iterative): {0}".format(ll)
from linked_list import randlinkedlist
def solution(node):
if not node or not node.next:
return
next = node.next
node.value = next.value
node.next = next.next
if __name__ == "__main__":
def _print(head):
cur = head
while cur:
print cur.value
cur = cur.next
ll = randlinkedlist()
print "Original linked list: "
_print(ll.head)
node = ll.head
for i in range(5):
node = node.next
solution(node)
print "New linked list: "
_print(ll.head)
#!/usr/bin/python
"""
Determine whether a linked list contains a loop as quickly as possible without using
any extra storage. Also, identify the location of the loop.
http://nbl.cewit.stonybrook.edu/algowiki/index.php/Data-structures-TADM2E-2
"""
from linked_list import randlinkedlist
ll = randlinkedlist(size=20)
print str(ll)
def find_cycle(llist):
slow, fast = llist.head, llist.head
while fast:
if fast.next is None:
break
if fast.next == slow:
return True
slow, fast = slow.next, fast.next.next
return False
print find_cycle(ll) # should return False
cur = ll.head
cycle_start = None
for i in range(ll.count - 1):
if i == 4:
cycle_start = cur
def solution3(head):
'''
Same as solution2.
'''
cur = head
while cur:
prev, runner = cur, cur.next
while runner:
if runner.value != cur.value:
prev.next = runner
prev = runner
runner = runner.next
prev.next = None
cur = cur.next
if __name__ == "__main__":
def _print(head):
cur = head
while cur:
print cur.value
cur = cur.next
ll = randlinkedlist(max=10)
print "Original linked list: "
_print(ll.head)
solution(ll.head)
print "Cleaned up linked list: "
_print(ll.head)
def solution3(head):
'''
Same as solution2.
'''
cur = head
while cur:
prev, runner = cur, cur.next
while runner:
if runner.value != cur.value:
prev.next = runner
prev = runner
runner = runner.next
prev.next = None
cur = cur.next
if __name__ == "__main__":
def _print(head):
cur = head
while cur:
print cur.value
cur = cur.next
ll = randlinkedlist(max=10)
print "Original linked list: "
_print(ll.head)
solution(ll.head)
print "Cleaned up linked list: "
_print(ll.head)
# Now that the linked lists are of equal size, add them
head, carry = add_forward(head1, head2)
if carry != 0:
node = Node(carry)
node.next, head = head, node # insert at front
return head
if __name__ == "__main__":
def _print(head):
cur = head
while cur:
print cur.value
cur = cur.next
ll1 = randlinkedlist(3, max=9)
ll2 = randlinkedlist(5, max=9)
print "Linked list 1:"
_print(ll1.head)
print "Linked list 2:"
_print(ll2.head)
print "Sum of linked lists (reversed):"
_print(solution1(ll1.head, ll2.head))
print "Sum of linked lists (not reversed):"
_print(solution2(ll1.head, ll2.head))
head2 = pad_with_zeros(head2, len1-len2)
# Now that the linked lists are of equal size, add them
head, carry = add_forward(head1, head2)
if carry != 0:
node = Node(carry)
node.next, head = head, node # insert at front
return head
if __name__ == "__main__":
def _print(head):
cur = head
while cur:
print cur.value
cur = cur.next
ll1 = randlinkedlist(3, max=9)
ll2 = randlinkedlist(5, max=9)
print "Linked list 1:"
_print(ll1.head)
print "Linked list 2:"
_print(ll2.head)
print "Sum of linked lists (reversed):"
_print(solution1(ll1.head, ll2.head))
print "Sum of linked lists (not reversed):"
_print(solution2(ll1.head, ll2.head))