{"id":5625,"date":"2019-09-29T21:17:49","date_gmt":"2019-09-30T04:17:49","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5625"},"modified":"2019-09-29T21:21:05","modified_gmt":"2019-09-30T04:21:05","slug":"leetcode-89-gray-code","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-89-gray-code\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 89. Gray Code"},"content":{"rendered":"\n

The gray code is a binary numeral system where two successive values differ in only one bit.<\/p>\n\n\n\n

Given a non-negative integer n<\/em> representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> 2\nOutput:<\/strong> [0,1,3,2]<\/code>\nExplanation:<\/strong>\n00 - 0\n01 - 1\n11 - 3\n10 - 2\n\nFor a given n<\/em>, a gray code sequence may not be uniquely defined.\nFor example, [0,2,3,1] is also a valid gray code sequence.\n\n00 - 0\n10 - 2\n11 - 3\n01 - 1\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> 0\nOutput:<\/strong> [0]\nExplanation:<\/strong> We define the gray code sequence to begin with 0.\n             A gray code sequence of n<\/em> has size = 2n<\/sup>, which for n<\/em> = 0 the size is 20<\/sup> = 1.\n             Therefore, for n<\/em> = 0 the gray code sequence is [0].<\/code><\/pre>\n\n\n\n
Solution: DP<\/strong><\/h2>\n\n\n\n
dp[0] = 0
dp[i] = dp[i – 1] + [x | 1 << (i – 1) for x in reversed(dp[i – 1])]<\/p>\n\n\n\n
Time complexity: O(2^n)
Space complexity: O(2^n)<\/p>\n\n\n\n
\n
C++<\/h2>\n
\n\n
\n\/\/ Author: Huahua\nclass Solution {\npublic:\n  vector grayCode(int n) {\n    vector> dp(n + 1);\n    dp[0] = {0};    \n    for (int i = 1; i <= n; ++i) {\n      dp[i] = dp[i - 1];\n      for (int j = dp[i - 1].size() - 1; j >= 0; --j)\n        dp[i].push_back(dp[i - 1][j] | (1 << (i - 1)));\n    }\n    return dp[n];\n  }\n};\n<\/pre>\n\n<\/div>
Python3<\/h2>\n
\n\n
\n# Author: Huahua\nclass Solution:\n  def grayCode(self, n: int) -> List[int]:    \n    dp = [0]\n    for i in range(n):      \n      dp = dp + [x | 1 << i for x in reversed(dp)]\n    return dp\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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