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Home/ Questions/Q 8005391
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T17:13:05+00:00 2026-06-04T17:13:05+00:00

Is there any general approach if one wanted to provide an immutable version of

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Is there any general approach if one wanted to provide an immutable version of e.g. LinkidList, implemented using as a linked sequence of nodes? I understand that in the case of ArrayList you would copy the underlying array, but in this case this is not that obvious to me…

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  1. Editorial Team
    Editorial Team
    2026-06-04T17:13:07+00:00Added an answer on June 4, 2026 at 5:13 pm

    Immutable lists are basically represented the same way as regular linked lists, except that all operations that would normally modify the list return a new one instead. This new list does not neccessarily need to contain a copy of the entire previous list but can reuse elements of it.

    I recommend implementing the following operations in the following ways:

    • Popping the element at the front: simply return a pointer to the next node. Complexity: O(1).
    • Pushing an element to the front: Create a new node that point to the first node of the old list and return it. O(1).
    • Concatenating list a with list b: copy the entire list a and let the pointer in the final node point to the beginning of list b. Note that this is faster than the same operation on mutable lists. O(length(a)).
    • Inserting at position x: Copy everything up to x, add a node with the new element to the back of the copy, and let that node point to the old list at position x + 1. O(x).
    • Removing the element at position x: practically the same as inserting. O(x).
    • Sorting: you can just use plain quick- or mergesort. It’s not much faster or slower than it would be on mutable lists. The only difference is that you can’t sort in place but will have to sort to a copy. O(n*log n).
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