你不需要使用正则表达式，你可以使用两个替换来解决你的问题：

You dont need to use regex, you can use two replace to solve your problem :

String replace = convert.replace("1", "one ").replace("0", "zero ");

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示例：

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Example :

int i = 55;
System.out.println(Integer.toBinaryString(i));
System.out.println(Integer.toBinaryString(i).replace("1", "one ").replace("0", "zero "));

输出

110111
one one zero one one one 

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超过一年后编辑。

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Edit after more than one year.

作为 @ Soheil Pourbafrani 在评论中询问，是否可以只遍历字符串一次，是的，你可以，但你需要使用这样的循环：

As @Soheil Pourbafrani ask in comment, is that possible to traverse the string only one time, yes you can, but you need to use a loop like so :

int i = 55;
char[] zerosOnes = Integer.toBinaryString(i).toCharArray();
String result = "";
for (char c : zerosOnes) {
    if (c == '1') {
        result += "one ";
    } else {
        result += "zero ";
    }
}
System.out.println(result);
=>one one two one one one

Java 8 +

如果您使用的是Java 8+，则更容易使用：

Java 8+

Or more easier if you are using Java 8+ you can use :

int i = 55;
String result = Integer.toBinaryString(i).chars()
        .mapToObj(c -> (char) c == '1' ? "one" : "two")
        .collect(Collectors.joining(" "));
=>one one two one one one