这是我迄今为止发现的运行最快的代码。我在23ms内运行了109015个长度为32的字节数组。我在VM上运行它，所以它在裸机上可能会运行得更快。

public static final char[] HEX_DIGITS =         {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

public static char[] encodeHex( final byte[] data ){
    final int l = data.length;
    final char[] out = new char[l<<1];
    for( int i=0,j=0; i<l; i++ ){
        out[j++] = HEX_DIGITS[(0xF0 & data[i]) >>> 4];
        out[j++] = HEX_DIGITS[0x0F & data[i]];
    }
    return out;
}

然后你就可以做了

String s = new String( encodeHex(myByteArray) );

如果你使用奇妙仙子，那么有:

package com.google.crypto.tink.subtle;

public final class Hex {
  public static String encode(final byte[] bytes) { ... }
  public static byte[] decode(String hex) { ... }
}

所以像这样的东西应该是有用的:

import com.google.crypto.tink.subtle.Hex;

byte[] bytes = {-1, 0, 1, 2, 3 };
String enc = Hex.encode(bytes);
byte[] dec = Hex.decode(enc)

其他人已经报道了一般情况。但是如果你有一个已知形式的字节数组，例如MAC地址，那么你可以:

byte[] mac = { (byte)0x00, (byte)0x00, (byte)0x00, (byte)0x00, (byte)0x00 };

String str = String.format("%02X:%02X:%02X:%02X:%02X:%02X",
                           mac[0], mac[1], mac[2], mac[3], mac[4], mac[5]); 

Use

Integer.toHexString((int)b);

这是你的快速方法:

    private static final String[] hexes = new String[]{
        "00","01","02","03","04","05","06","07","08","09","0A","0B","0C","0D","0E","0F",
        "10","11","12","13","14","15","16","17","18","19","1A","1B","1C","1D","1E","1F",
        "20","21","22","23","24","25","26","27","28","29","2A","2B","2C","2D","2E","2F",
        "30","31","32","33","34","35","36","37","38","39","3A","3B","3C","3D","3E","3F",
        "40","41","42","43","44","45","46","47","48","49","4A","4B","4C","4D","4E","4F",
        "50","51","52","53","54","55","56","57","58","59","5A","5B","5C","5D","5E","5F",
        "60","61","62","63","64","65","66","67","68","69","6A","6B","6C","6D","6E","6F",
        "70","71","72","73","74","75","76","77","78","79","7A","7B","7C","7D","7E","7F",
        "80","81","82","83","84","85","86","87","88","89","8A","8B","8C","8D","8E","8F",
        "90","91","92","93","94","95","96","97","98","99","9A","9B","9C","9D","9E","9F",
        "A0","A1","A2","A3","A4","A5","A6","A7","A8","A9","AA","AB","AC","AD","AE","AF",
        "B0","B1","B2","B3","B4","B5","B6","B7","B8","B9","BA","BB","BC","BD","BE","BF",
        "C0","C1","C2","C3","C4","C5","C6","C7","C8","C9","CA","CB","CC","CD","CE","CF",
        "D0","D1","D2","D3","D4","D5","D6","D7","D8","D9","DA","DB","DC","DD","DE","DF",
        "E0","E1","E2","E3","E4","E5","E6","E7","E8","E9","EA","EB","EC","ED","EE","EF",
        "F0","F1","F2","F3","F4","F5","F6","F7","F8","F9","FA","FB","FC","FD","FE","FF"
    };

    public static String byteToHex(byte b) {
        return hexes[b&0xFF];
    }

只是添加我的两分，因为我看到许多答案使用字符数组和/或使用StringBuilder实例，并声称是快速或更快。

由于我使用ASCII表组织有一个不同的想法，代码点48-57为0-9，代码点65-70为a-f，代码点97-102为a-f，我想测试哪个想法是最快的。

因为我已经好几年没有做过类似的事情了，所以我要做一个广泛的拍摄。我在不同大小的数组中使用了10亿字节(1M, 1K, 10)，因此每个数组有1000倍1M字节，每个数组有1M倍1000字节，每个数组有100M倍10字节。

结果是1-F中的char数组胜出。使用char数组而不是StringBuilder作为输出也很容易(对象更少，不需要测试容量，在增长时不需要新数组或复制)。此外，当使用foreach (for(var b: bytes))循环时，你似乎会得到一个小的惩罚。

使用我的想法的版本大约是。每个数组1M字节时慢15%，每个数组1000字节时慢21%，每个数组10字节时慢18%。StringBuilder版本分别慢了210%、380%和310%。

这很糟糕，但也不是那么出乎意料，因为在第一级缓存中查找小数组胜过if和add... .(一个缓存访问+偏移量计算vs.一个if，一个跳转，一个add ->不确定跳转thou)。

我的版本:

public static String bytesToHex(byte [] bytes) {
    
    char [] result = new char [bytes.length * 2];
    
    for(int index = 0; index < bytes.length; index++) {
        int v = bytes[index]; 

        int upper = (v >>> 4) & 0xF;
        result[index * 2] = (char)(upper + (upper < 10 ? 48 : 65 - 10));
        
        int lower = v & 0xF;
        result[index * 2 + 1] = (char)(lower + (lower < 10 ? 48 : 65 - 10));
    }
    
    return new String(result);
}

PS:是的，我做了多次测试，每次都做了最好的测试，做了热身，也做了100亿字符的测试，以确保相同的图片... .