Arduino as Socks5 Proxy Server

Still WIP because the buffer manipulation is so f**king problematic.

Hardware: Arduino Uno + W5100 Ethernet Module

Tested working through curl.


LeetCode 403. Frog Jump

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.

Given a list of stones’ positions (in units) in sorted ascending order, determine if the frog is able to cross the river by landing on the last stone. Initially, the frog is on the first stone and assume the first jump must be 1 unit.

If the frog’s last jump was k units, then its next jump must be either k – 1, k, or k + 1 units. Note that the frog can only jump in the forward direction.

思路:这个题目可以看作是动态规划。青蛙的一个状态可以表示为{当前的k,当前的位置(可以用stone的索引表示)},最初的时候状态就是{ 1, 0 }。在某一个状态下,青蛙可以尝试向前移动{k-1, k, k+1}步,如果能成功跳到那就进入了下一个状态。要注意的是这种状态的遍历是有重复的,所以需要一个东西记录已经尝试过(当然是尝试失败的,不然直接return true)的状态们,不然很容易超时。



LeetCode 89. Gray Code

Divide and conquer!

Add a leading 0 or 1 bit to your existing results so here comes two sub-results: 0{xxxxx} and 1{xxxxx}. How to merge them? Just keep the first sub group and append the second group items reversely so that there is only a 1-bit difference between the end of group 0 and “start” of group 2.


LeetCode 402. Remove K Digits

如果直接用DFS做的话轻轻松松就会stack overflow。


贪心法之所以能用是因为我们去尽量保证了a1<a2< …< x< y,如果这些个数字里要选一个删掉的话,肯定是删掉y带来的结果最好。


LeetCode 269. Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:

Example 2:

Example 3:


  1. You may assume all letters are in lowercase.
  2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
  3. If the order is invalid, return an empty string.
  4. There may be multiple valid order of letters, return any one of them is fine.



The solution is to build a directed map and find the topological ordering of the characters. Once the map is built, we can always find the “terminal” character which has no succeeding, i.e. the out-degree of that node would be zero. The we remove the character (the node and its edges) from the map and try to find the next such character, so on so forth.

For example,  the map built will be:



LeetCode 611. Valid Triangle Number


时间复杂度是O(n*log(n) + n^2) = O(n^2)



LeetCode 523. Continuous Subarray Sum

这道题跟 其实是很像的,但本题的特点就是corner case非常多…


LeetCode 23. Merge k Sorted Lists

Use max heap, the time complexity is O( log(k) * N )


LeetCode 222. Count Complete Tree Nodes