网站综合信息 www.codesays.com
- 标题:
- Code Says | C code. C code run. Run co
- 关键字:
- 描述:
- 域名信息
- 域名年龄:12年2个月28天 注册日期:2012年10月13日 到期时间:2021年10月13日
邮箱:3e31f51b29de63ab7bba7c35f 电话:+1.877-237-6466
注册商:GOOGLE INC.
- 服务器空间
- IP:该域名无法解析为IP
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- 备案号:
域名流量Alexa排名
域名注册Whois信息
codesays.com
域名年龄: 12年2个月28天注册时间: 2012-10-13
到期时间: 2021-10-13
注 册 商: GOOGLE INC.
注册邮箱: 3e31f51b29de63ab7bba7c35f
联系电话: +1.877-237-6466
获取时间: 2016年02月17日 14:35:31
Domain Name: CODESAYS.COM
Registrar: GOOGLE INC.
Sponsoring Registrar IANA ID: 895
Whois Server: whois.rrpproxy.net
Referral URL: http://domains.google.com
Name Server: NS-CLOUD-D1.GOOGLEDOMAINS.COM
Name Server: NS-CLOUD-D2.GOOGLEDOMAINS.COM
Name Server: NS-CLOUD-D3.GOOGLEDOMAINS.COM
Name Server: NS-CLOUD-D4.GOOGLEDOMAINS.COM
Status: clientTransferProhibited https://www.icann.org/epp#clientTransferProhibited
Updated Date: 2016-02-01
Creation Date: 2012-10-13
Expiration Date: 2021-10-13
>>> Last update of whois database: Wed, 2016-Feb-17 06:38:55 GMT <<<
For more information on Whois status codes, please visit https://icann.org/epp
Domain Name: codesays.com
Registry Domain ID: 1751888302_DOMAIN_COM-VRSN
Registrar WHOIS Server: whois.rrpproxy.net
Registrar URL: https://domains.google.com/
Updated Date: 2016-02-01T07:27:57.0Z
Creation Date: 2012-10-13T16:21:54.0Z
Registrar Registration Expiration Date: 2021-10-13T16:21:54.0Z
Registrar: Google Inc.
Registrar IANA ID: 895
Registrar Abuse Contact Email: registrar-abuse[at]google.com
Registrar Abuse Contact Phone: +1.877-237-6466
Domain Status: clientTransferProhibited https://icann.org/epp#clientTransferProhibited
Registry Registrant ID: Not Available From Registry
Registrant Name: On behalf of codesays.com OWNER
Registrant Organization: c/o whoisproxy.com Ltd.
Registrant Street: Plaza Level,41 Shortland Street
Registrant City: Auckland
Registrant State/Province:
Registrant Postal Code: 1010
Registrant Country: NZ
Registrant Phone: +64.48319528
Registrant Phone Ext:
Registrant Fax:
Registrant Fax Ext:
Registrant Email: 7151f9a1983147171f1321cf9955b52fac1898a3e31f51b29de63ab7bba7c35forg
Registry Admin ID: Not Available From Registry
Admin Name: On behalf of codesays.com ADMIN
Admin Organization: c/o whoisproxy.com Ltd.
Admin Street: Plaza Level,41 Shortland Street
Admin City: Auckland
Admin State/Province:
Admin Postal Code: 1010
Admin Country: NZ
Admin Phone: +64.48319528
Admin Phone Ext:
Admin Fax:
Admin Fax Ext:
Admin Email: 7151f9a1983147171f1321cf9955b52fac1898a3e31f51b29de63ab7bba7c35forg
Registry Tech ID: Not Available From Registry
Tech Name: On behalf of codesays.com TECH
Tech Organization: c/o whoisproxy.com Ltd.
Tech Street: Plaza Level,41 Shortland Street
Tech City: Auckland
Tech Postal Code: 1010
Tech State/Province:
Tech Country: NZ
Tech Phone: +64.48319528
Tech Phone Ext:
Tech Fax:
Tech Fax Ext:
Tech Email: 7151f9a1983147171f1321cf9955b52fac1898a3e31f51b29de63ab7bba7c35forg
Name Server: ns-cloud-d1.googledomains.com
Name Server: ns-cloud-d2.googledomains.com
Name Server: ns-cloud-d3.googledomains.com
Name Server: ns-cloud-d4.googledomains.com
DNSSEC: unsigned
Whoisprivacy: 1
URL of the ICANN WHOIS Data Problem Reporting System: https://wdprs.internic.net/
>>> Last update of WHOIS database: 2016-02-17T06:39:09.0Z <<<
Registry Billing ID: Not Available From Registry
Billing Name: On behalf of codesays.com BILLING
Billing Organization: c/o whoisproxy.com Ltd.
Billing Street: Plaza Level,41 Shortland Street
Billing City: Auckland
Billing State/Province:
Billing Postal Code: 1010
Billing Country: NZ
Billing Phone: +64.48319528
Billing Phone Ext:
Billing Fax:
Billing Fax Ext:
Billing Email: 7151f9a1983147171f1321cf9955b52fac1898a3e31f51b29de63ab7bba7c35forg
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网站头文件(HTTP Header)
HTTP/1.1 200 OK 服务器:nginx 访问时间:2016年02月17日 14:39:06 类型:text/html Transfer-Encoding: chunked 连接:关闭 动作:Accept-Encoding 过期时间:2016年02月17日 14:39:05 缓存控制:不缓存 Content-Encoding: gzip
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抓取时间:2019年09月13日 15:17:16
网址:http://www.codesays.com/
标题:Code Says | C code. C code run. Run code run…please!
关键字:
描述:
主体:
网址:http://www.codesays.com/
标题:Code Says | C code. C code run. Run code run…please!
关键字:
描述:
主体:
Code SaysC code. C code run. Run code run…please!HomeCracking the Coding InterviewCodilityJobdu OJLeetCodeAbout MeStackOverflow ProfileLinkedIn ProfileContact MeSolution to Find Peak Element by LeetCode December 26, 2014 Sheng LeetCode, Python, 0 Question: https://oj.leetcode.com/problems/find-peak-element/ Question Name: Find Peak Element The key point is: return the index to ANY ONE of the peaks. So O(logN) is possible. Solution to Find Peak Element by LeetCodePythonclass Solution:# @param num, a list of integer# @base, the base index for tail recursion.# @return an integerdef findPeakElement(self, num, base = 0):# Refer to www.geeksforgeeks.org/find-a-peak-in-a-given-array/# Because num[-1] = num[n] = negative infinity, if there is only# one element in the list, it is a peak.if len(num) == 1: return base# If there are two two elements in the list, the greater one is# the peak.if len(num) == 2:if num[0] > num[1]: return baseelse: return base + 1mid = (len(num) - 1) // 2if num[mid-1] < num[mid] and num[mid+1] < num[mid]:# The middle element is a peak.return mid + baseelif num[mid] < num[mid-1]:# There must be one or more peak(s) in the left part.return self.findPeakElement(num[:mid], base)else:# There must be one or more peak(s) in the right part.return self.findPeakElement(num[mid+1:], mid + 1 + base)123456789101112131415161718192021222324252627class Solution: # @param num, a list of integer # @base, the base index for tail recursion. # @return an integer def findPeakElement(self, num, base = 0): # Refer to www.geeksforgeeks.org/find-a-peak-in-a-given-array/ # Because num[-1] = num[n] = negative infinity, if there is only # one element in the list, it is a peak. if len(num) == 1: return base # If there are two two elements in the list, the greater one is # the peak. if len(num) == 2: if num[0] > num[1]: return base else: return base + 1 mid = (len(num) - 1) // 2 if num[mid-1] < num[mid] and num[mid+1] < num[mid]: # The middle element is a peak. return mid + base elif num[mid] < num[mid-1]: # There must be one or more peak(s) in the left part. return self.findPeakElement(num[:mid], base) else: # There must be one or more peak(s) in the right part. return self.findPeakElement(num[mid+1:], mid + 1 + base)Solution to Intersection of Two Linked Lists by LeetCode December 25, 2014 Sheng LeetCode, Python, 0 Question: https://oj.leetcode.com/problems/intersection-of-two-linked-lists/ Question Name: Intersection of Two Linked Lists Solution to Intersection of Two Linked Lists by LeetCodePython# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution:# Get the length of the given linked list# @param ListNo