leetcode: 19. Remove Nth Node From End of List

# Given a linked list, remove the nth node from the end of list and return its head.
#
# For example,
#
# Given linked list: 1->2->3->4->5, and n = 2.
#
# After removing the second node from the end, the linked list becomes 1->2->3->5.
#
# Note:
# Given n will always be valid.
# Try to do this in one pass.

class ListNode():
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution():
    def removeNthFromEnd(self, head, n):
        slow = fast = dummy = ListNode(0)
        dummy.next = head
        while n:
            fast, n = fast.next, n-1
        while fast.next:
            slow, fast = slow.next, fast.next
        slow.next = slow.next.next
        return dummy.next


if __name__ == "__main__":
    head, head.next, head.next.next, head.next.next.next, head.next.next.next.next \
        = ListNode(1), ListNode(2), ListNode(3), ListNode(4), ListNode(5)
    result = Solution().removeNthFromEnd(head, 2)
    assert '{0}->{1}->{2}->{3}'.format(result.val, result.next.val, result.next.next.val, result.next.next.next.val) \
           == '1->2->3->5'