{
   "question_id": 89,
   "question_title": "格雷编码",
   "difficulty": "中等",
   "question_content": "<p>格雷编码是一个二进制数字系统，在该系统中，两个连续的数值仅有一个位数的差异。</p>\n<p>给定一个代表编码总位数的非负整数<em> n</em>，打印其格雷编码序列。即使有多个不同答案，你也只需要返回其中一种。</p>\n<p>格雷编码序列必须以 0 开头。</p>\n<p>&nbsp;</p>\n<p><strong>示例 1:</strong></p>\n<pre><strong>输入:</strong>&nbsp;2<strong><br />输出:</strong>&nbsp;[0,1,3,2]<strong><br />解释:</strong>00 - 001 - 111 - 310 - 2对于给定的&nbsp;<em>n</em>，其格雷编码序列并不唯一。例如，[0,2,3,1]&nbsp;也是一个有效的格雷编码序列。00 - 010 - 211 - 301 - 1</pre>\n<p><strong>示例&nbsp;2:</strong></p>\n<pre><strong>输入:</strong>&nbsp;0<strong><br />输出:</strong>&nbsp;[0]<strong><br />解释:</strong> 我们定义格雷编码序列必须以 0 开头。给定编码总位数为 <em>n</em> 的格雷编码序列，其长度为 2<sup>n</sup>。当 <em>n</em> = 0 时，长度为 2<sup>0</sup> = 1。因此，当 <em>n</em> = 0 时，其格雷编码序列为 [0]。</pre>",
   "topic_link": "https://bbs.csdn.net/topics/600470797",
   "cpp": "#include <stdio.h>\n#include <stdlib.h>\nint *grayCode(int n, int *returnSize)\n{\n\tif (n < 0)\n\t{\n\t\treturn NULL;\n\t}\n\tint i, count = 1 << n;\n\tint *codes = malloc(count * sizeof(int));\n\tfor (i = 0; i < count; i++)\n\t{\n\t\tcodes[i] = (i >> 1) ^ i;\n\t}\n\t*returnSize = 1 << n;\n\treturn codes;\n}\nint main(int argc, char **argv)\n{\n\tif (argc != 2)\n\t{\n\t\tfprintf(stderr, \"Usage: ./test n\\n\");\n\t\texit(-1);\n\t}\n\tint i, count;\n\tint *list = grayCode(atoi(argv[1]), &count);\n\tfor (i = 0; i < count; i++)\n\t{\n\t\tprintf(\"%d \", list[i]);\n\t}\n\tprintf(\"\\n\");\n\treturn 0;\n}",
   "java": "class Solution {\n\tpublic List<Integer> grayCode(int n) {\n\t\tList<Integer> res = new ArrayList<>();\n\t\tres.add(0);\n\t\tint cur;\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tint change = 1 << i;\n\t\t\tcur = res.size() - 1;\n\t\t\twhile (cur >= 0) {\n\t\t\t\tres.add(res.get(cur) ^ change);\n\t\t\t\tcur--;\n\t\t\t}\n\t\t}\n\t\treturn res;\n\t}\n}",
   "js": "",
   "python": "class Solution(object):\n\tdef grayCode(self, n):\n\t\t\"\"\"\n\t\t:type n: int\n\t\t:rtype: List[int]\n\t\t\"\"\"\n\t\tres = [0]\n\t\tfor i in range(n):\n\t\t\tfor j in reversed(range(len(res))):\n\t\t\t\tres.append(res[j] + (1 << i))\n\t\treturn res\nif __name__ == \"__main__\":\n\ts = Solution()\n\tprint (s.grayCode(2))",
   "status": 1,
   "keywords": "位运算,数学,回溯",
   "license": {
      "cpp": "https://github.com/begeekmyfriend/leetcode",
      "python": "https://github.com/qiyuangong/leetcode",
      "java": "https://blog.csdn.net/qq_25406563/article/details/85065072"
   },
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