Feedback #1
Feedback #1
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| for (int i = 0; i < index; i++) { // O(n) | ||
| item = item.next; // O(1) | ||
| } | ||
| return item.data; // O(1) |
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Generally, great implementations of these operations, but consider an edge case we're not actively testing for: what happens if index exceeds the bounds of the linked list?
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i think i fixed that with the commit
| LinkedList reversedList = new LinkedList(); // O(1) | ||
| LinkedList.Node item = list.head; // O(1) | ||
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| for (int i = 0; i < list.length(); i++) { // O(n) |
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While at first blush, this seemingly looks like an O(n) implementation, what is the time complexity of list.length()? In relation to this for loop's iterations, how often would length() be calculated and what would the resulting complexity be? What do you think about improving our efficiency here for larger list lengths?
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changed it so that length is set to variable before
| item = item.next; // O(1) | ||
| } | ||
| return reversedList; // O(1) | ||
| } // Algorithmic Complexity = O(n) |
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Nice use of prepend()!
Bonus Credit: Curious to get your thoughts on if you see an implementation that allows you to reverse the list in place (i.e. consume no added memory), but perhaps manipulates the original list. The API is not precise enough to indicate what the semantic behaviour should strictly be, but you made the safe assumption given the a list is specified with a return value suggesting that this should not be a destructive operation. Though, as an exercise, do you want to give an in-place implementation a shot just within the Github comments here?
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done i think? I commited
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| assertEquals(2, list.head.data); | ||
| assertEquals(3, list.head.next.data); | ||
| } |
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🎉 🥇
I appreciate you adding more tests!
| item = item.next; // O(1) | ||
| } | ||
| if (item.next.next == null) { // O(1) | ||
| item.next = null; // O(1) |
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I'm curious about the item.next.next == null condition; could you explain it to me? Is this intended to serve as a final statement for the case index == 0 and would this strictly be applicable in cases where the list is longer than three elements?
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if you want to remove the last thing on a linked list there is no item.next.next so you just set the last thing on the linked list to be null and so it becomes the tail. Otherwise if the index is in the middle area, then i will skip over one link and delete that one. That is my thought process Im not sure if it is right
…n its own list not make a new one
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