algorithms/191-number-of-1-bits.ipynb

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    "### [位1的个数](https://leetcode-cn.com/problems/number-of-1-bits/)\n",
    "\n",
    "请实现一个函数，输入一个整数（以二进制串形式），输出该数二进制表示中 1 的个数。例如，把 9 表示成二进制是 1001，有 2 位是 1。因此，如果输入 9，则该函数输出 2。\n",
    "\n",
    "#### 示例 1：\n",
    "```\n",
    "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",
    "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",
    "输入：11111111111111111111111111111101\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",
    "### 提示：\n",
    "> 输入必须是长度为 32 的 二进制串 。\n",
    "\n"
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    "def hammingWeight(n: int) -> int:\n",
    "    ans = 0\n",
    "    bin_n = bin(n)\n",
    "    for i in range(2, len(bin_n)):\n",
    "        if bin_n[i] == '1':\n",
    "            ans += 1\n",
    "    return ans"
   ]
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    "hammingWeight(9)"
   ]
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    "def hammingWeight(n: int) -> int:\n",
    "    ans = 0\n",
    "    for i in range(32):\n",
    "        if n & 1 == 1:\n",
    "            ans += 1\n",
    "        n = n >> 1\n",
    "    return ans"
   ]
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     "execution_count": 27,
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    "hammingWeight(9)"
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   "source": [
    "hammingWeight(-2147483648)"
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