Sockets - Evelynn #25
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| # Time Complexity: O(1) | ||
| # Space Complexity: O(1) | ||
| def add_first(value) | ||
| if self.length != 0 |
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You can do this without calling the length method since it doesn't matter how long the list is to add an element to the front.
Also since you're doing .length, this causes your algorithm to be O(n) time complexity.
| def find_min | ||
| raise NotImplementedError | ||
| current = @head | ||
| (self.length - 1).times do |
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Since .length is a method and it traverses the list, it would be good to save that result in a variable rather than calling the method multiple times.
| end | ||
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| middle = self.length / 2 | ||
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| middle_value = self.get_at_index(middle) |
| # Space Complexity | ||
| def add_last(value) | ||
| raise NotImplementedError | ||
| index_of_n = self.length - 1 - n |
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Why not use the get_at_index method again?
| # Time Complexity: O(n) where n is the number of nodes in the list | ||
| # Space Complexity: O(1) | ||
| def has_cycle | ||
| if self.length <= 1 |
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Just a point, if it has a cycle, then the length method will get stuck in an infinite loop.
| if a == b | ||
| return true | ||
| else | ||
| a = a.next |
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This won't work if the cycle is more than 2 nodes long.
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