在java中将十进制转换为格雷码 [英] Convert decimal to gray code in java

问题描述

最近出现了一个问题：编写算法将十进制数转换为n位格雷码。

Had a question come up recently which was: write the algorithm to convert a decimal number to an n-bit gray code.

例如：
使用1位（最简单）：

So for example: Using 1-bit (simplest):

0 -> 0
1 -> 1

使用2位

0 -> 00
1 -> 01
2 -> 11
3 -> 10

使用3位

0 -> 000
1 -> 001
2 -> 011
3 -> 010
4 -> 110
5 -> 111
6 -> 101
7 -> 100

推荐答案

写下以下内容并认为我会分享因为我没有看到很多Java实现出现在这里：

Wrote the following and figured I'd share it as I don't see many Java implementations showing up on here:

static String getGreyCode(int myNum, int numOfBits) {
    if (numOfBits == 1) {
        return String.valueOf(myNum);
    }

    if (myNum >= Math.pow(2, (numOfBits - 1))) {
        return "1" + getGreyCode((int)(Math.pow(2, (numOfBits))) - myNum - 1, numOfBits - 1);
    } else {
        return "0" + getGreyCode(myNum, numOfBits - 1);
    }
}

static String getGreyCode(int myNum) {

    //Use the minimal bits required to show this number
    int numOfBits = (int)(Math.log(myNum) / Math.log(2)) + 1;
    return getGreyCode(myNum, numOfBits);
}

要测试此项，您可以通过以下任一方式调用它：

And to test this, you can call it in either of the following ways:

System.out.println("Grey code for " + 7 + " at n-bit: " + getGreyCode(7));
System.out.println("Grey code for " + 7 + " at 5-bit: " + getGreyCode(7, 5));

或者循环遍历第i位的所有灰色代码组合：

Or loop through all the possible combinations of grey codes up to the ith-bit:

for (int i = 1; i <= 4; i++) {
        for (int j = 0; j < Math.pow(2, i); j++)
            System.out.println("Grey code for " + j + " at " + i + "-bit: " + getGreyCode(j, i));

希望这对人们有帮助！

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