c ++ language divide and conquer strategy to generate Gray code

Gray code is a string of 2 ^ n sequences. No sequence identical elements, each of length n is (0,1) strings, only just adjacent to a different element.

Here are a few low Golay code

A Gray code (2 ^ 1 = 2)	2-bit Gray code (2 ^ 2 = 4)	3-bit Gray code (2 ^ 3 = 8)	4-bit Gray code (2 ^ 4 = 16)	other…
0	00	000	0000	…
1	01	001	0001
	11	011	0011
	10	010	0010
		110	0110
		111	0111
		101	0101
		100	0100
			1100
			1101
			1111
			1110
			1010
			1011
			1001
			1000

Not difficult to draw from the above table:

• A Gray code has two code words 0
• (N + 1) before 2 ^ n-bit Gray code codewords is equal to n-bit code words of a Gray code, in the order written, prefix 0
• (N + 1) in the Gray code bit 2 ^ n codewords is equal to n-bit code words of a Gray code, written in reverse order, prefix 1
• n + 1-bit Gray code set = n-bit Gray code set (order) prefixed 0 + n-bit Gray code set (reverse) prefixed 1

Use two-dimensional array and recursively to solve the problem:

• n = 1, arr[0][0] = 0, arr[1][0] = 1
• n> 1, the recursive illustrated as follows:

	0	1	2	3	4	…	n
0	0	0	0
1	1	0	0
2	1	1	0
3	0	1	0
4	0	1	1
5	1	1	1
6	1	0	1
7	0	0	1
…
n

Gray code output, the output from right to left

#include <iostream> 
#include<math.h> 
using namespace std; 
void Gray(int **arr,int sum, int n) {     //Gray生成函数
    if (n == 1) {
        arr[0][0] = 0;
        arr[1][0] = 1;
        return;
    }
    Gray(arr,sum / 2, n - 1);
    for (int i = 0; i < sum / 2; i++) {		//循环作用为添加 0 和 1 
        arr[i][n - 1] = 0;
        arr[sum - i - 1][n - 1] = 1;
    }	
    for (int i = sum / 2; i < sum; i++) {	//循环作用为 倒序复制
        for (int j = 0; j < n - 1; j++)
            arr[i][j] = arr[sum - i - 1][j];
    }
}
int main()
{
    int n;
    int** arr;
    cout << "输入n: ";
    cin >> n;
    int sum = pow(2, n);  
    arr = new int* [sum];       //动态二维数组
    for (int i = 0; i < sum; i++)
        arr[i] = new int[n];
    Gray(arr, sum, n);		//生成格雷码
    for (int i = 0; i < sum; i++) {	//输出，方向为右向左  ←
        for (int j = n-1; j>=0  ; j--)
        {
            cout << arr[i][j];
        }
        cout << endl;
    }
    for (int i = 0; i < sum; i++)	//释放内存
        delete[] arr[i];
    delete[] arr;
    return 0;
}

Run renderings

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