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leetcode Bulb Switcher

There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the _n_th round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Example:

Given n = 3.

At first, the three bulbs are [off, off, off].

After first round, the three bulbs are [on, on, on].

After second round, the three bulbs are [on, off, on].

After third round, the three bulbs are [on, off, off].

So you should return 1, because there is only one bulb is on.

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leetcode Maximum Product of Word Lengths

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.

Example 1:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"] Return 16 The two words can be "abcw", "xtfn".

Example 2:

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"] Return 4 The two words can be "ab", "cd".

Example 3:

Given ["a", "aa", "aaa", "aaaa"] Return 0 No such pair of words.

Follow up: Could you do better than O(_n_2), where n is the number of words?

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前言：

当数据量大的时候，一台机器可能不足以满足性能的需求，如存储空间、内存不足，而垂直扩展十分昂贵，故用分片进行水平扩展是一个很好的解决方法。

本文介绍了mongodb 进行分片的方法和步骤

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Given an Iterator class interface with methods: next() and hasNext(), design and implement a PeekingIterator that support the peek() operation -- it essentially peek() at the element that will be returned by the next call to next()

Here is an example. Assume that the iterator is initialized to the beginning of the list: [1, 2, 3].

Call next() gets you 1, the first element in the list.

Now you call peek() and it returns 2, the next element. Calling next() after that still return 2.

You call next() the final time and it returns 3, the last element. Calling hasNext() after that should return false.

Hint:

  1. Think of "looking ahead". You want to cache the next element.
  2. Is one variable sufficient? Why or why not?
  3. Test your design with call order of peek() before next() vs next() before peek().
  4. For a clean implementation, check out Google's guava library source code.

Follow up: How would you extend your design to be generic and work with all types, not just integer?

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leetcode Remove Duplicate Letters

Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.

Example:

Given "bcabc" Return "abc"

Given "cbacdcbc" Return "acdb"

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