CODEWITHSUNDEEP
pythonlinked-list
zeyuxie
zeyuxie

Reputation: 75

linked list function that return a list

    class LN:
        def __init__(self,value,next=None):
            self.value = value
            self.next  = next

    def add_after(ll,value,new):
        item = ll
        while item is not None:
            if item.value == value:
                newnode = LN(new, item.next)
                item.next = newnode
                break
            else:
                item = item.next

my add_after function takes a list such as [2,1,8,2,2,4,2,5,2], and a value and a new value. it add the new value to every value that appears in the list. for example:

l = [2,1,8,2,2,4,2,5,2], calling add_after(a,2,-1) returns 2->-1->1->8->2->-1->2->-1->4->2->-1->5->2->-1->None. it add -1 after every 2 in the list

the problem with my add_after function is that it only add the new value to the first value that appears in the list.

for example, if I call add_after(ll,2,-1) to the list [1,8,2,4,2,5,2], it returns 1->8->2->-1->4->2->5->2->None

can someone help me to fix my add_after function so that it can apply to all of the value in the list? many thanks.

Upvotes: 1

Views: 101

Answers (2)

quantummind
quantummind

Reputation: 2136

The break statement breaks the loop, so it will not got through the rest of the list.

There were other problems as well. A corrected version of the code:

class TN:
    def __init__(self,value,next=None):
        self.value = value
        self.next = next

    def add_after(self, value, new):
        item = self
        while item is not None:
            if item.value == value:
                newnode = TN(new, item.next)
                item.next = newnode
                item = newnode.next
            else:
                item = item.next

Upvotes: 1

Patrick Haugh
Patrick Haugh

Reputation: 61032

Below is a quick way to turn a list into a doubly linked list. Read through and understand it, then you should be able to modify it to suit your needs.

def link_list(l):
    if not l:
        return None
    head = TN(l[0])
    prev = head
    for val in l[1:]:
        curr = TN(val, prev)
        prev.right = curr
        prev=curr
    return head

Upvotes: 1

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