Subtract first and last nodes of linked list

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Problem from www.interviewbit.com.




Problem : Given a singly linked list, modify the value of first half nodes such that :



1st node’s new value = the last node’s value - first node’s current value



2nd node’s new value = the second last node’s value - 2nd node’s current value,
and so on …



NOTE :
If the length L of linked list is odd, then the first half implies at first floor(L/2) nodes. So, if L = 5, the first half refers to first 2 nodes.
If the length L of linked list is even, then the first half implies at first L/2 nodes. So, if L = 4, the first half refers to first 2 nodes.
Example :



Given linked list 1 -> 2 -> 3 -> 4 -> 5,
Return 4 -> 2 -> 3 -> 4 -> 5



as
for first node, 5 - 1 = 4
for second node, 4 - 2 = 2



Try to solve the problem using constant extra space.




This is my solution on the website's shell. How can this code be better?



# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution:
# @param A : head node of linked list
# @return the head node in the linked list
def subtract(self, A):

if not A.next:
return A
fast = A
slow = A
prev = None
cnt = 0
# reach till half
while fast and fast.next:
fast = fast.next.next
slow = slow.next
cnt += 1
if fast:
start = slow.next
else:
start = slow
#reverse the latter half
while start:
temp = start.next
start.next = prev
prev = start
start = temp

firstA = A
lastA = prev
prev1 = prev
currA = A

# modify values of first half
while cnt:
currA.val = lastA.val - firstA.val
firstA = firstA.next
lastA = prev.next

prev = prev.next

currA = currA.next
cnt -= 1


# reverse the list again
new_prev =None
while prev1:
temp = prev1.next
prev1.next = new_prev
new_prev = prev1
prev1 = temp

return A






share|improve this question

























    up vote
    0
    down vote

    favorite












    Problem from www.interviewbit.com.




    Problem : Given a singly linked list, modify the value of first half nodes such that :



    1st node’s new value = the last node’s value - first node’s current value



    2nd node’s new value = the second last node’s value - 2nd node’s current value,
    and so on …



    NOTE :
    If the length L of linked list is odd, then the first half implies at first floor(L/2) nodes. So, if L = 5, the first half refers to first 2 nodes.
    If the length L of linked list is even, then the first half implies at first L/2 nodes. So, if L = 4, the first half refers to first 2 nodes.
    Example :



    Given linked list 1 -> 2 -> 3 -> 4 -> 5,
    Return 4 -> 2 -> 3 -> 4 -> 5



    as
    for first node, 5 - 1 = 4
    for second node, 4 - 2 = 2



    Try to solve the problem using constant extra space.




    This is my solution on the website's shell. How can this code be better?



    # Definition for singly-linked list.
    # class ListNode:
    # def __init__(self, x):
    # self.val = x
    # self.next = None

    class Solution:
    # @param A : head node of linked list
    # @return the head node in the linked list
    def subtract(self, A):

    if not A.next:
    return A
    fast = A
    slow = A
    prev = None
    cnt = 0
    # reach till half
    while fast and fast.next:
    fast = fast.next.next
    slow = slow.next
    cnt += 1
    if fast:
    start = slow.next
    else:
    start = slow
    #reverse the latter half
    while start:
    temp = start.next
    start.next = prev
    prev = start
    start = temp

    firstA = A
    lastA = prev
    prev1 = prev
    currA = A

    # modify values of first half
    while cnt:
    currA.val = lastA.val - firstA.val
    firstA = firstA.next
    lastA = prev.next

    prev = prev.next

    currA = currA.next
    cnt -= 1


    # reverse the list again
    new_prev =None
    while prev1:
    temp = prev1.next
    prev1.next = new_prev
    new_prev = prev1
    prev1 = temp

    return A






    share|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Problem from www.interviewbit.com.




      Problem : Given a singly linked list, modify the value of first half nodes such that :



      1st node’s new value = the last node’s value - first node’s current value



      2nd node’s new value = the second last node’s value - 2nd node’s current value,
      and so on …



      NOTE :
      If the length L of linked list is odd, then the first half implies at first floor(L/2) nodes. So, if L = 5, the first half refers to first 2 nodes.
      If the length L of linked list is even, then the first half implies at first L/2 nodes. So, if L = 4, the first half refers to first 2 nodes.
      Example :



      Given linked list 1 -> 2 -> 3 -> 4 -> 5,
      Return 4 -> 2 -> 3 -> 4 -> 5



      as
      for first node, 5 - 1 = 4
      for second node, 4 - 2 = 2



      Try to solve the problem using constant extra space.




      This is my solution on the website's shell. How can this code be better?



      # Definition for singly-linked list.
      # class ListNode:
      # def __init__(self, x):
      # self.val = x
      # self.next = None

      class Solution:
      # @param A : head node of linked list
      # @return the head node in the linked list
      def subtract(self, A):

      if not A.next:
      return A
      fast = A
      slow = A
      prev = None
      cnt = 0
      # reach till half
      while fast and fast.next:
      fast = fast.next.next
      slow = slow.next
      cnt += 1
      if fast:
      start = slow.next
      else:
      start = slow
      #reverse the latter half
      while start:
      temp = start.next
      start.next = prev
      prev = start
      start = temp

      firstA = A
      lastA = prev
      prev1 = prev
      currA = A

      # modify values of first half
      while cnt:
      currA.val = lastA.val - firstA.val
      firstA = firstA.next
      lastA = prev.next

      prev = prev.next

      currA = currA.next
      cnt -= 1


      # reverse the list again
      new_prev =None
      while prev1:
      temp = prev1.next
      prev1.next = new_prev
      new_prev = prev1
      prev1 = temp

      return A






      share|improve this question











      Problem from www.interviewbit.com.




      Problem : Given a singly linked list, modify the value of first half nodes such that :



      1st node’s new value = the last node’s value - first node’s current value



      2nd node’s new value = the second last node’s value - 2nd node’s current value,
      and so on …



      NOTE :
      If the length L of linked list is odd, then the first half implies at first floor(L/2) nodes. So, if L = 5, the first half refers to first 2 nodes.
      If the length L of linked list is even, then the first half implies at first L/2 nodes. So, if L = 4, the first half refers to first 2 nodes.
      Example :



      Given linked list 1 -> 2 -> 3 -> 4 -> 5,
      Return 4 -> 2 -> 3 -> 4 -> 5



      as
      for first node, 5 - 1 = 4
      for second node, 4 - 2 = 2



      Try to solve the problem using constant extra space.




      This is my solution on the website's shell. How can this code be better?



      # Definition for singly-linked list.
      # class ListNode:
      # def __init__(self, x):
      # self.val = x
      # self.next = None

      class Solution:
      # @param A : head node of linked list
      # @return the head node in the linked list
      def subtract(self, A):

      if not A.next:
      return A
      fast = A
      slow = A
      prev = None
      cnt = 0
      # reach till half
      while fast and fast.next:
      fast = fast.next.next
      slow = slow.next
      cnt += 1
      if fast:
      start = slow.next
      else:
      start = slow
      #reverse the latter half
      while start:
      temp = start.next
      start.next = prev
      prev = start
      start = temp

      firstA = A
      lastA = prev
      prev1 = prev
      currA = A

      # modify values of first half
      while cnt:
      currA.val = lastA.val - firstA.val
      firstA = firstA.next
      lastA = prev.next

      prev = prev.next

      currA = currA.next
      cnt -= 1


      # reverse the list again
      new_prev =None
      while prev1:
      temp = prev1.next
      prev1.next = new_prev
      new_prev = prev1
      prev1 = temp

      return A








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      asked Jun 9 at 18:09









      Latika Agarwal

      861216




      861216




















          1 Answer
          1






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          up vote
          1
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          More functions, please.



          Every time you feel compelled to comment a block of code, consider factoring it out into a properly named function. For example,



          def subtract(A):
          start = reach_till_half(A)
          prev = reverse(start)
          modify(A, prev)
          reverse(prev)


          Of course names could (and should) be better.






          share|improve this answer





















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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote













            More functions, please.



            Every time you feel compelled to comment a block of code, consider factoring it out into a properly named function. For example,



            def subtract(A):
            start = reach_till_half(A)
            prev = reverse(start)
            modify(A, prev)
            reverse(prev)


            Of course names could (and should) be better.






            share|improve this answer

























              up vote
              1
              down vote













              More functions, please.



              Every time you feel compelled to comment a block of code, consider factoring it out into a properly named function. For example,



              def subtract(A):
              start = reach_till_half(A)
              prev = reverse(start)
              modify(A, prev)
              reverse(prev)


              Of course names could (and should) be better.






              share|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                More functions, please.



                Every time you feel compelled to comment a block of code, consider factoring it out into a properly named function. For example,



                def subtract(A):
                start = reach_till_half(A)
                prev = reverse(start)
                modify(A, prev)
                reverse(prev)


                Of course names could (and should) be better.






                share|improve this answer













                More functions, please.



                Every time you feel compelled to comment a block of code, consider factoring it out into a properly named function. For example,



                def subtract(A):
                start = reach_till_half(A)
                prev = reverse(start)
                modify(A, prev)
                reverse(prev)


                Of course names could (and should) be better.







                share|improve this answer













                share|improve this answer



                share|improve this answer











                answered Jun 9 at 18:38









                vnp

                36.4k12890




                36.4k12890






















                     

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