Created
April 24, 2010 04:35
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And *that* is how to reverse a linked list. I'm an idiot for messing this up.
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| class Node(object): | |
| def __init__(self, value): | |
| self.value = value | |
| self.next = None | |
| def __str__(self): | |
| """Return a string representation of the list starting here.""" | |
| values = [] | |
| curr = self | |
| while curr is not None: | |
| print repr(curr) | |
| values.append(str(curr.value)) | |
| curr = curr.next | |
| return ",".join(values) | |
| def reverse(head): | |
| """reverse a linked list in-place and return the new head.""" | |
| curr = head.next | |
| prev = head | |
| prev.next = None # Break the link between the head and whatever's after it. | |
| while curr is not None: | |
| # Retain a reference to whatever curr points to now. | |
| tmp = curr.next | |
| # make curr point at the thing before it. | |
| curr.next = prev | |
| # move everybody forward. | |
| prev = curr | |
| curr = tmp | |
| # when this loop terminates curr should be None but prev | |
| # will be the last element, now the first. | |
| return prev | |
| if __name__ == "__main__": | |
| # Some demo code. | |
| a = Node(4) | |
| a.next = Node(5) | |
| a.next.next = Node(6) | |
| a.next.next.next = Node(7) | |
| print a | |
| b = reverse(a) | |
| print b |
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