摘要：
Great learning for me:https://www.hackerrank.com/rest/contests/master/challenges/lucky-numbers/hackers/turuthok/download_solution Basically it is memo 阅读全文
posted @ 2016-04-07 02:40
Tonix
阅读(313)
评论(0)
推荐(0)
摘要：
Besides the common stack-based solutions, here is another recursive NestedIterator solution :) And here's my stack-based solution: 阅读全文
posted @ 2016-04-06 03:18
Tonix
阅读(270)
评论(0)
推荐(0)
摘要：
A simple variation to "Longest Substring with At Most Two Distinct Characters". A typical sliding window problem. 阅读全文
posted @ 2016-04-04 02:53
Tonix
阅读(197)
评论(0)
推荐(0)
摘要：
Nothing special. Just take care of corner cases. class Solution { public: /** * @param head a ListNode * @oaram v1 an integer * @param v2 an integer * 阅读全文
posted @ 2016-04-01 13:37
Tonix
阅读(176)
评论(0)
推荐(0)
摘要：
An intuitive DFS one. 阅读全文
posted @ 2016-03-31 00:38
Tonix
阅读(135)
评论(0)
推荐(0)
摘要：
Neat DP problem to go. If you don't like log(), you can use lookup table. And, this is an amazing solution: https://leetcode.com/discuss/93501/simple- 阅读全文
posted @ 2016-03-21 12:15
Tonix
阅读(137)
评论(0)
推荐(0)
摘要：
Nice one. Try to learn the way of thinking, with the help of the below image: https://leetcode.com/discuss/89336/best-submission-searching-for-the-cro 阅读全文
posted @ 2016-03-13 15:02
Tonix
阅读(169)
评论(0)
推荐(0)
摘要：
Recursion of course. But please make sure you cover all the cases. class Solution(object): # return: max value with node picked, max val w\o node pick 阅读全文
posted @ 2016-03-12 14:09
Tonix
阅读(139)
评论(0)
推荐(0)
摘要：
Backtrace idea is straightforward. But there are always smarter solutions - like this: https://leetcode.com/discuss/87314/non-recursive-time-space-sol 阅读全文
posted @ 2016-03-12 08:15
Tonix
阅读(136)
评论(0)
推荐(0)
摘要：
My first reaction was to use Manacher - which is O(n*k*k) - n is the no. of words, and k is the average length of words. Code of Manacher based soluti 阅读全文
posted @ 2016-03-10 14:57
Tonix
阅读(222)
评论(0)
推荐(0)
摘要：
First problem to learn Max Flow. Ford-Fulkerson is a group of algorithms - Dinic is one of it.It is an iterative process: we use BFS to check augament 阅读全文
posted @ 2016-02-17 12:23
Tonix
阅读(372)
评论(0)
推荐(0)
摘要：
Here is the mind flow: My intuition was to use Fenwick tree - that works but that's O(nlgn). That's too generalized and the magic number is 3 - that m 阅读全文
posted @ 2016-02-16 15:14
Tonix
阅读(135)
评论(0)
推荐(0)
摘要：
A typical recursion use: class Solution(object): max_size = 0 # return: isValid, minVal, maxVal, nodeCnt def go(self, root): if root.left is None and 阅读全文
posted @ 2016-02-12 14:31
Tonix
阅读(274)
评论(0)
推荐(0)
摘要：
Intuitive one to learn about Grundy basic :) Now every pile becomes a game, so we need to use Sprague-Grundy Theory. Calculation is quite intuitive - 阅读全文
posted @ 2016-02-11 09:04
Tonix
阅读(310)
评论(0)
推荐(0)
摘要：
Please note: VROOK cannot go back-ward - that leads to a simple variation to Game of Nim: just XOR. t = int(input()) for _ in range(t): n = int(input( 阅读全文
posted @ 2016-02-11 05:15
Tonix
阅读(229)
评论(0)
推荐(0)
摘要：
A variation to 0-1 Knapsack. (No Python code got fully AC. Python is too slow for this problem) #include <cmath> #include <cstdio> #include <vector> # 阅读全文
posted @ 2016-02-03 14:32
Tonix
阅读(414)
评论(0)
推荐(0)
摘要：
https://leetcode.com/discuss/77133/o-n-o-1-after-median-virtual-indexing Please note the "interleaved" version of partial.. nice idea.. 阅读全文
posted @ 2016-02-03 12:36
Tonix
阅读(125)
评论(0)
推荐(0)
摘要：
Yes a typical stack based problem. Please take care of corner cases. class Solution { public: bool isValidSerialization(string preorder) { // Get toke 阅读全文
posted @ 2016-02-03 06:08
Tonix
阅读(260)
评论(0)
推荐(0)
摘要：
I got lost totally - it is a whole range of numbers! Actually, we can check each number within the range and pick it greedily. https://leetcode.com/di 阅读全文
posted @ 2016-02-01 05:49
Tonix
阅读(136)
评论(0)
推荐(0)
摘要：
This is 'Difficult' - I worked out it within 45mins, and unlocked HackerRank Algorithm Level 80 yeah! So the idea is straight forward: 1. sort the inp 阅读全文
posted @ 2016-01-28 14:16
Tonix
阅读(318)
评论(0)
推荐(0)
浙公网安备 33010602011771号