柳婼 の blog 我不管反正我最萌
LeetCode 461. Hamming Distance. The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Input x 1, y 4. 1 0 0 0 1. 4 0 1 0 0. The above arrows point to positions where the corresponding bits are different. 对于右移一位采用x 2和 y 2的方式对于比较最后一位即比较x 2 和 y 2统计不相同的次数cnt直到x和y都等于0为止. Class Solution public int hammingDistanceint x, int y int cnt 0; whilex! 0 ifx 2!
OVERVIEW
The website liuchuo.net presently has an average traffic ranking of zero (the lower the more traffic). We have crawled twenty pages within the web page liuchuo.net and found one website linking to liuchuo.net.
Pages Parsed
20Links to this site
1LIUCHUO.NET TRAFFIC
The website liuchuo.net is seeing diverging quantities of traffic in the past the year.
Date Range
1 week
1 month
3 months
This Year
Last Year
All time
Date Range
All time
This Year
Last Year
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LINKS TO LIUCHUO.NET
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WHAT DOES LIUCHUO.NET LOOK LIKE?
LIUCHUO.NET SERVER
I diagnosed that a single page on liuchuo.net took three thousand two hundred and fifty-six milliseconds to load. We could not observe a SSL certificate, so our crawlers consider liuchuo.net not secure.
Load time
3.256 sec
SSL
NOT SECURE
IP
121.199.1.155
FAVICON
SERVER SOFTWARE
We identified that this domain is using the Apache operating system.SITE TITLE
柳婼 の blog 我不管反正我最萌DESCRIPTION
LeetCode 461. Hamming Distance. The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Input x 1, y 4. 1 0 0 0 1. 4 0 1 0 0. The above arrows point to positions where the corresponding bits are different. 对于右移一位采用x 2和 y 2的方式对于比较最后一位即比较x 2 和 y 2统计不相同的次数cnt直到x和y都等于0为止. Class Solution public int hammingDistanceint x, int y int cnt 0; whilex! 0 ifx 2!PARSED CONTENT
The website has the following on the web site, "The Hamming distance between two integers is the number of positions at which the corresponding bits are different." I analyzed that the web page also said " Given two integers x and y, calculate the Hamming distance." They also stated " Input x 1, y 4. 1 0 0 0 1. 4 0 1 0 0. The above arrows point to positions where the corresponding bits are different. 对于右移一位采用x 2和 y 2的方式对于比较最后一位即比较x 2 和 y 2统计不相同的次数cnt直到x和y都等于0为止. Class Solution public int hammingDistanceint x, int y int cnt 0; whilex! 0 ifx 2!."SEE OTHER WEB PAGES
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