2017ๅนด4ๆœˆ19ๆ—ฅๆ˜ŸๆœŸไธ‰

Quiz 8-- Princeton Algorithm

Java autoboxing and equals(). Consider two ๐š๐š˜๐šž๐š‹๐š•๐šŽ values ๐šŠ and ๐š‹ and their corresponding <tt>Double</tt> values ๐šก and ๐šข.
  • Find values such that (๐šŠ==๐š‹) is ๐š๐š›๐šž๐šŽ but ๐šก.๐šŽ๐šš๐šž๐šŠ๐š•๐šœ(๐šข) is ๐š๐šŠ๐š•๐šœ๐šŽ.
  • Find values such that (๐šŠ==๐š‹) is ๐š๐šŠ๐š•๐šœ๐šŽ but ๐šก.๐šŽ๐šš๐šž๐šŠ๐š•๐šœ(๐šข) is ๐š๐š›๐šž๐šŽ.
Hint: IEEE floating point arithmetic has some peculiar rules for ๐Ÿถ.๐Ÿถ๐Ÿถ.๐Ÿถ, and ๐™ฝ๐šŠ๐™ฝ. Java requires that ๐šŽ๐šš๐šž๐šŠ๐š•๐šœ() implements an equivalence relation.


0.0 == -0.0        Double.valueOf(0.0).equals(Double.valueOf(-0.0))
Double.NaN == Double.NaN   Double.valueOf(Double.NaN).equals(Double.valueOf(Double.NaN))




Check if a binary tree is a BST. Given a binary tree where each ๐™ฝ๐š˜๐š๐šŽ contains a key, determine whether it is a binary search tree. Use extra space proportional to the height of the tree.


Hint: design a recursive function ๐š’๐šœ๐™ฑ๐š‚๐šƒ(๐™ฝ๐š˜๐š๐šŽ๐šก,๐™บ๐šŽ๐šข๐š–๐š’๐š—,๐™บ๐šŽ๐šข๐š–๐šŠ๐šก) that determines whether ๐šก is the root of a binary search tree with all keys between ๐š–๐š’๐š— and ๐š–๐šŠ๐šก.



  if (Nodex == null) return true;
  if ( Nodex.value < Keymin || Nodex.value > KeyMax) return false;

  return isBST(Nodex.left, Keymin, Nodex.value) && isBST(Nodex.right, Nodex.value , Keymax)


Inorder traversal with constant extra space. Design an algorithm to perform an inorder traversal of a binary search tree using only a constant amount of extra space.

Hint: you may modify the BST during the traversal provided you restore it upon completion.



Web tracking. Suppose that you are tracking n web sites and m users and you want to support the following API:
  • User visits a website.
  • How many times has a given user visited a given site?
What data structure or data structures would you use?

Hint: maintain a symbol table of symbol tables.





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