def getReversedList(head_in):
list_new = LinkedList()
curr = head_in
while curr:
list_new.insert(curr.val, 0)
curr = curr.next
return list_new.getHeadNode()
## Tests For Singly Linked List Implementation
from myllist import LinkedList
list1 = LinkedList()
print "Size is: " + str(list1.size())
list1.append(1)
list1.append(2)
list1.append(3)
print "Size is: " + str(list1.size())
for idx in range(list1.size()):
print list1.get(idx)
list1.insert(4, 3)
print "Size is: " + str(list1.size())
for idx in range(list1.size()):
print list1.get(idx)
list1.remove(3)
print "Size is: " + str(list1.size())
for idx in range(list1.size()):
print list1.get(idx)
## If we had a doubly linked list (with a tail pointer), we could have done this in O(n/2)
import sys
sys.path.append('./classes')
from myllist import LinkedList
def getReversedList(head_in):
list_new = LinkedList()
curr = head_in
while curr:
list_new.insert(curr.val, 0)
curr = curr.next
return list_new.getHeadNode()
def isPalindrome(head1):
head2 = getReversedList(head1)
while head1:
if head1.val != head2.val:
return False
head1 = head1.next
head2 = head2.next
return True
list1 = LinkedList()
for n in "":
list1.append(n)
print isPalindrome(list1.getHeadNode())
return head1.val * multiplier + head2.val * multiplier + sumForwardHelper(
head1.next, head2.next, multiplier / 10)
def findMaxMultiplier(head1, head2):
length1 = 0
curr = head1
while curr:
length1 += 1
curr = curr.next
length2 = 0
curr = head2
while curr:
length2 += 1
curr = curr.next
return pow(10, max(length1, length2) - 1)
list1 = LinkedList()
for n in [7, 1, 6]:
list1.append(n)
list2 = LinkedList()
for n in [5, 9, 2]:
list2.append(n)
headNode1 = list1.getHeadNode()
headNode2 = list2.getHeadNode()
print sumReverse(headNode1, headNode2)
print sumForward(headNode1, headNode2)
## CTCI - Linked Lists Question 3
## Delete middle element, given only access to that element
import sys
sys.path.append('./classes')
from myllist import LinkedList
def deleteMiddleElmt(mnode):
if mnode is None or mnode.next is None:
return
curr = mnode
nextElmt = mnode.next
while nextElmt.next:
tmp = curr.next
curr.val = nextElmt.val
curr = tmp
nextElmt = tmp.next
curr.val = nextElmt.val
curr.next = None
list1 = LinkedList()
for n in [1, 2, 3, 4, 5]:
list1.append(n)
middleNode = list1.getNode(3)
list1.printList()
deleteMiddleElmt(middleNode)
list1.printList()
def returnLoopNode(head):
if head is None:
return None
itemSet = Set()
runner = head
itemSet.add(head)
while runner.next:
if runner.next in itemSet:
return runner.next
itemSet.add(runner.next)
runner = runner.next
return None
list1 = LinkedList()
for n in [1, 2, 3, 4, 5, 6]:
list1.append(n)
endNode = list1.getNode(5)
loopNode = list1.getNode(2)
endNode.next = loopNode
headNode = list1.getHeadNode()
result = returnLoopNode(headNode)
if result:
print result.val
else:
print result
def partition(head,x): if head is None: return searcher = head partitioner = head while partitioner and partitioner.val <= x: partitioner = partitioner.next searcher = searcher.next while searcher: if searcher.val > x: searcher = searcher.next else: smallVal = searcher.val searcher.val = partitioner.val partitioner.val = smallVal partitioner = partitioner.next searcher = searcher.next list1 = LinkedList() for n in [6,2,3,4,5,7]: list1.append(n) headNode = list1.getHeadNode() list1.printList() partition(headNode,3) list1.printList()
## CTCI - Linked Lists Question 2
## Return kth last element of the list (k=0 returns last element)
## Assumes we know the size of the list
## Assumes we want the value of the element, not the node itself
import sys
sys.path.append('./classes')
from myllist import LinkedList
def returnKthToLast(inlist, k):
totalSize = inlist.size()
if totalSize > 0:
stop = totalSize - 1 - k
return returnKthToLastHelper(inlist.getHeadNode(), stop, 0)
return None
def returnKthToLastHelper(curr, stop, pos):
if pos == stop:
return curr.val
return returnKthToLastHelper(curr.next, stop, pos + 1)
list1 = LinkedList()
for n in [1, 2, 3, 4, 5, 6]:
list1.append(n)
for n in [0, 1, 2, 3, 4, 5]:
print returnKthToLast(list1, n)