<?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>LeetCode 191 on Jeanphilo Blog</title><link>https://shio-chan-dev.github.io/jeanblog/zh/tags/leetcode-191/</link><description>Recent content in LeetCode 191 on Jeanphilo Blog</description><generator>Hugo -- 0.165.0</generator><language>zh-cn</language><lastBuildDate>Wed, 15 Jul 2026 00:00:00 +0800</lastBuildDate><atom:link href="https://shio-chan-dev.github.io/jeanblog/zh/tags/leetcode-191/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://shio-chan-dev.github.io/jeanblog/zh/alg/leetcode/191-number-of-1-bits/</link><pubDate>Wed, 15 Jul 2026 00:00:00 +0800</pubDate><guid>https://shio-chan-dev.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></channel></rss>