<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:content="http://purl.org/rss/1.0/modules/content/"><channel><title>二进制 on Jeanphilo Blog</title><link>https://jeanphilogong.github.io/jeanblog/zh/tags/%E4%BA%8C%E8%BF%9B%E5%88%B6/</link><description>Recent content in 二进制 on Jeanphilo Blog</description><generator>Hugo -- 0.168.0</generator><language>zh-cn</language><lastBuildDate>Wed, 15 Jul 2026 00:00:00 +0800</lastBuildDate><atom:link href="https://jeanphilogong.github.io/jeanblog/zh/tags/%E4%BA%8C%E8%BF%9B%E5%88%B6/index.xml" rel="self" type="application/rss+xml"/><item><title>LeetCode 191：位 1 的个数，如何跳过无关的零位</title><link>https://jeanphilogong.github.io/jeanblog/zh/alg/leetcode/191-number-of-1-bits/</link><pubDate>Wed, 15 Jul 2026 00:00:00 +0800</pubDate><guid>https://jeanphilogong.github.io/jeanblog/zh/alg/leetcode/191-number-of-1-bits/</guid><description>从逐位右移 baseline 出发，推导 n &amp;amp; (n - 1) 每次清除最低位的一个 1，并用循环 invariant 证明 LeetCode 191。</description></item><item><title>LeetCode 338：比特位计数，如何复用较小数字的结果</title><link>https://jeanphilogong.github.io/jeanblog/zh/alg/leetcode/hot100/338-counting-bits/</link><pubDate>Wed, 15 Jul 2026 00:00:00 +0800</pubDate><guid>https://jeanphilogong.github.io/jeanblog/zh/alg/leetcode/hot100/338-counting-bits/</guid><description>从对 0..n 每个数字重复统计的 baseline 出发，推导 answer[i] = answer[i &amp;amp; (i - 1)] + 1，在线性时间内完成比特位计数。</description></item><item><title>判断一个数是否为 2 的幂（Power of Two）：位运算 O(1) ACERS 解析（LeetCode 231）</title><link>https://jeanphilogong.github.io/jeanblog/zh/alg/leetcode/231-power-of-two/</link><pubDate>Wed, 21 Jan 2026 13:22:53 +0800</pubDate><guid>https://jeanphilogong.github.io/jeanblog/zh/alg/leetcode/231-power-of-two/</guid><description>用位运算 O(1) 判断整数是否为 2 的幂，含工程场景、常见误区与多语言实现。</description></item></channel></rss>