刷题技巧.ipynb

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   "source": [
    "计算一个数的二进制位中的最低一位1（lowbit），可以用`x & -x`实现。"
   ]
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   "source": [
    "lowbit = 5 & (-5)\n",
    "lowbit"
   ]
  },
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    "异或运算有以下三个性质。\n",
    "\n",
    "1. 任何数和 0 做异或运算，结果仍然是原来的数，即 $a⊕0=a$。\n",
    "2. 任何数和其自身做异或运算，结果是 0，即 $a⊕a=0$。\n",
    "3. 异或运算满足交换律和结合律，即 $a⊕b⊕a=b⊕a⊕a=b⊕(a⊕a)=b⊕0=b$\n",
    "\n",
    "出题，给你一个 非空 整数数组 nums ，除了某个元素只出现一次以外，其余每个元素均出现两次。找出那个只出现了一次的元素。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from functools import reduce\n",
    "class Solution:\n",
    "    def singleNumber(self, nums: list) -> int:\n",
    "        return reduce(lambda x, y: x ^ y, nums)"
   ]
  }
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