C code. C code run. Run code run… please!
This little blog is served on Google Cloud. Since the last day of October, it was stopped by my mistake and I cannot restart the service. Google Cloud always complains: Operation type [start] failed with message “The zone ‘projects/???/zones/us-west2-b’ does … Read More »
Security is always one of the top concerns about websites, including WordPress. And login security is the foundation of many other security features, and usually the first line of defense. Aside from using vulnerabilities in WordPress, a less-skilled but much … Read More »
Sometimes, the visitors posted their code in the comment section. That is an awesome way to share and communicate. However, in some cases, partial of the code inside pre/code section is missing. This unexpected feature bothered me for a long … Read More »
Question: http://oj.leetcode.com/problems/3sum/ Question Name: 3Sum Classic question. For more information, you could read the Wikipedia.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 | class Solution: # @return num[i]list of lists of length 3, [[val1,val2,val3]] def threeSum(self, num): num.sort() result = [] i = 0 # For the first item while i < len(num) - 2: j = i + 1 # For the middle item k = len(num) - 1 # For the last item while j < k: if num[i]+ num[j] + num[j] > 0 or num[i]+ num[k]+ num[k]< 0: # num[k] >= any num[j], num[j] <= any num[k] # Impossible to find a answer in the future break if num[i]+ num[j]+num[k]== 0: # Because the num is sorted, so the num[i] <= num[j] <= num[k] # And in every round, i or j/k is different from the previous # round. Therefore, the answer [num[i], num[j], num[k]] is new # and unique for the result set. result.append([num[i], num[j], num[k]]) # Skip duplicate num[j-1] and num[k+1] j += 1 while j < k+1 and num[j] == num[j-1]: j += 1 k -= 1 while k > j-1 and num[k] == num[k+1]: k -= 1 elif num[i] + num[j]+ num[k]< 0: # Skip duplicate num[j-1] j += 1 while j < k+1 and num[j] == num[j-1]: j += 1 else: # Skip duplicate num[k+1] k -= 1 while k > j-1 and num[k] == num[k+1]: k -= 1 # Skip duplicate num[i-1] i += 1 while i < len(num)-1 and num[i] == num[i-1]: i += 1 return result |
AppFog is a cloud platform, supporting PHP, Ruby, Java, and so on. It is a good and free solution for small website, like personal blogs. My blog (codesays.com) is currently hosted on AppFog. But AppFog does not support persistent file … Read More »
A hello world program, in most cases, is the first program of newbies. As a newbie of blog, I am happy to begin my site with this hello world article, and a short piece of hello world program in Python3: … Read More »