Time - Alex #26
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Time - Alex #26
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CheezItMan
reviewed
Sep 1, 2020
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| if @head.nil? | ||
| @head = Node.new(value) | ||
| else | ||
| new_nodelet = Node.new(value, @head) | ||
| new_nodelet.next = @head | ||
| @head = new_nodelet | ||
| end |
There was a problem hiding this comment.
You could rewrite this as:
Suggested change
| if @head.nil? | |
| @head = Node.new(value) | |
| else | |
| new_nodelet = Node.new(value, @head) | |
| new_nodelet.next = @head | |
| @head = new_nodelet | |
| end | |
| @head = Node.new(value, @head) |
This way the new node's next points to the old head (which might be nil).
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def search(value) |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def find_max |
There was a problem hiding this comment.
Correct, but space complexity would be O(1) since you don't create a new data structure, but just track the greatest value found so far.
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def find_min |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def length |
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| return nil if @head.nil? | ||
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| current = @head | ||
| count = 0 | ||
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| until count == index | ||
| current = current.next | ||
| count += 1 | ||
| end | ||
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| return current.data |
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What if the length is less than index?
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def visit |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def delete(value) |
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What if you are deleting the 1st node?
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| last = @head | ||
| current = @head | ||
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| until current.data == value || current.nil? | ||
| last = current | ||
| current = current.next | ||
| end | ||
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| last.next = current.next |
There was a problem hiding this comment.
Suggested change
| last = @head | |
| current = @head | |
| until current.data == value || current.nil? | |
| last = current | |
| current = current.next | |
| end | |
| last.next = current.next | |
| if @head.data == value | |
| @head = @head.next | |
| return | |
| end | |
| last = nil | |
| current = @head | |
| until current.data == value || current.nil? | |
| last = current | |
| current = current.next | |
| end | |
| unless current.nil? | |
| last.next = current.next | |
| end |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def reverse |
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