Leetcode: Gray Code
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题目:
The gray code is a binary numeral system where two successive values differ in only one bit.
Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.
For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:
00 - 001 - 111 - 310 - 2
Note:
For a given n, a gray code sequence is not uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.
For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.
Gray Code:
在一组数的编码中,若任意两个相邻的代码只有一位二进制数不同,则称这种编码为格雷码(Gray Code)。格雷码当初是为了通信,现在则常用于模拟-数字转换和位置-数字转换中。
Gray Code转换方法:
递归生成码表
这种方法基于格雷码是反射码的事实,利用递归的如下规则来构造:
1. 1位格雷码有两个码字
2. n位格雷码中的前2(n-1)个码字等于n-1位格雷码的码字,按顺序书写,加前缀0
3. (n+1)位格雷码中的后2(n-1)个码字等于n-1位格雷码的码字,按逆序书写,加前缀1
C++参考代码(用时7ms):
class Solution{public: vector<int> grayCode(int n) { vector<int> codes; codes.push_back(0); for (int i = 0; i < n; ++i) { int one = 1 << i;//最高位的数字1 int size = int(codes.size()); //这个循环要倒序哦 for (int j = size - 1; j >= 0; --j) { codes.push_back(one + codes[j]); } } return codes; }};
异或转换
二进制码->格雷码(编码):
1. 从0到2n-1编号
2. 从最右边一位起,依次将每一位与左边一位异或(XOR),作为对应格雷码该位的值,最左边一位不变(相当于左边是0);
格雷码->二进制码(解码):
从左边第二位起,将每位与左边一位解码后的值异或,作为该位解码后的值(最左边一位依然不变)。
C++代码(用时9ms):
class Solution{public: vector<int> grayCode(int n) { vector<int> codes; int size = 1 << n;//相当于pow(2, n) for (int i = 0; i < size; ++i) { codes.push_back((i >> 1) ^ i); } return codes; }};
我发现把第十行的右移一位codes.push_back((i >> 1) ^ i)(用了9ms)改成除法运算codes.push_back((i / 2) ^ i)(用了6ms)运算还变快了!
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