{ "data": { "question": { "questionId": "100292", "questionFrontendId": "剑指 Offer 15", "categoryTitle": "LCOF", "boundTopicId": 86250, "title": "二进制中1的个数 LCOF", "titleSlug": "er-jin-zhi-zhong-1de-ge-shu-lcof", "content": "English description is not available for the problem. Please switch to Chinese.", "translatedTitle": "二进制中1的个数", "translatedContent": "

编写一个函数，输入是一个无符号整数（以二进制串的形式），返回其二进制表达式中数字位数为 '1' 的个数（也被称为 汉明重量).）。

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提示：

\n\n\n\n

 

\n\n

示例 1：

\n\n
\n输入：n = 11 (控制台输入 00000000000000000000000000001011)\n输出：3\n解释：输入的二进制串 00000000000000000000000000001011 中，共有三位为 '1'。\n
\n\n

示例 2：

\n\n
\n输入：n = 128 (控制台输入 00000000000000000000000010000000)\n输出：1\n解释：输入的二进制串 00000000000000000000000010000000 中，共有一位为 '1'。\n
\n\n

示例 3：

\n\n
\n输入：n = 4294967293 (控制台输入 11111111111111111111111111111101，部分语言中 n = -3）\n输出：31\n解释：输入的二进制串 11111111111111111111111111111101 中，共有 31 位为 '1'。
\n\n

 

\n\n

提示：

\n\n\n\n

 

\n\n

注意：本题与主站 191 题相同：https://leetcode-cn.com/problems/number-of-1-bits/

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