有没有更好的方法来插入“|” 转换成二进制字符串 rep 得到这个 10|000|001

Question

"|"给定十进制的二进制字符串表示形式，是否有更好的方法插入字符串200 = 11001000 此函数返回字符串 =11|001|000

虽然这个功能有效，但看起来很笨拙！！为什么

在 GO 中做一个简单的字符插入这么难？？？

func (i Binary) FString() string {

    a := strconv.FormatUint(i.Get(), 2)

    y := make([]string, len(a), len(a)*2)

    data := []rune(a)

    r := []rune{}

    for i := len(data) - 1; i >= 0; i-- {

        r = append(r, data[i])

    }

    for j := len(a) - 1; j >= 0; j-- {

        y = append(y, string(r[j]))

        if ((j)%3) == 0 && j > 0 {

            y = append(y, "|")

        }

    }

    return strings.Join(y, "")

}

Answer 1

取决于你叫什么更好。我会使用正则表达式。在这种情况下，复杂性来自于从右侧插入分隔符。如果我们填充字符串使其长度是 3 的倍数，我们可以从左侧插入分隔符。我们可以很容易地使用正则表达式|在每三个字符之前插入。然后，我们可以去掉前导|+ 填充。func (i Binary) FString() string {    a := strconv.FormatUint(i.Get(), 2)    pad_req := len(a) % 3    padding := strings.Repeat("0", (3 - pad_req))    a = padding + a    re := regexp.MustCompile("([01]{3})")    a = re.ReplaceAllString(a, "|$1")    start := len(padding) + 1    if len(padding) == 3 {        // If we padded with "000", we want to remove the `|` before *and* after it        start = 5    }    a = a[start:]    return a}

Answer 2

如果性能不重要并且您只想要一个紧凑的版本，您可以将输入数字复制到输出，并在将|一组 2 写入输出时插入一个符号。组从右到左计数，因此当从左到右复制数字时，第一组可能更小。因此，组内的数字计数器不一定从0第一组的情况开始，而是从len(input)%3.这是一个例子：func Format(s string) string {&nbsp; &nbsp; b, count := &bytes.Buffer{}, len(s)%3&nbsp; &nbsp; for i, r := range s {&nbsp; &nbsp; &nbsp; &nbsp; if i > 0 && count == i%3 {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; b.WriteRune('|')&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; b.WriteRune(r)&nbsp; &nbsp; }&nbsp; &nbsp; return b.String()}测试它：for i := uint64(0); i < 10; i++ {&nbsp; &nbsp; fmt.Println(Format(strconv.FormatUint(i, 2)))}fmt.Println(Format(strconv.FormatInt(1234, 2)))输出（在Go Playground上试试）：0110111001011101111|0001|00110|011|010|010如果您必须多次执行此操作并且性能确实很重要，请查看我对问题的回答：How to fmt.Printf an integer with many comma基于此，一个快速的解决方案可以是：func Format(s string) string {&nbsp; &nbsp; out := make([]byte, len(s)+(len(s)-1)/3)&nbsp; &nbsp; for i, j, k := len(s)-1, len(out)-1, 0; ; i, j = i-1, j-1 {&nbsp; &nbsp; &nbsp; &nbsp; out[j] = s[i]&nbsp; &nbsp; &nbsp; &nbsp; if i == 0 {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return string(out)&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; if k++; k == 3 {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; j, k = j-1, 0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; out[j] = '|'&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}输出当然是一样的。在Go Playground上尝试一下。

Answer 3

这是一个分区问题。您可以使用此功能：func partition(s, separator string, pLen int) string {&nbsp; &nbsp; if pLen < 1 || len(s) == 0 || len(separator) == 0 {&nbsp; &nbsp; &nbsp; &nbsp; return s&nbsp; &nbsp; }&nbsp; &nbsp; buffer := []rune(s)&nbsp; &nbsp; L := len(buffer)&nbsp; &nbsp; pCount := L / pLen&nbsp; &nbsp; result := []string{}&nbsp; &nbsp; index := 0&nbsp; &nbsp; for ; index < pCount; index++ {&nbsp; &nbsp; &nbsp; &nbsp; _from := L - (index+1)*pLen&nbsp; &nbsp; &nbsp; &nbsp; _to := L - index*pLen&nbsp; &nbsp; &nbsp; &nbsp; result = append(result, string(buffer[_from:_to]))&nbsp; &nbsp; }&nbsp; &nbsp; if L%pLen != 0 {&nbsp; &nbsp; &nbsp; &nbsp; result = append(result, string(buffer[0:L-index*pLen]))&nbsp; &nbsp; }&nbsp; &nbsp; for h, t := 0, len(result)-1; h < t; h, t = h+1, t-1 {&nbsp; &nbsp; &nbsp; &nbsp; result[t], result[h] = result[h], result[t]&nbsp; &nbsp; }&nbsp; &nbsp; return strings.Join(result, separator)}并且s := partition("11001000", "|", 3)会给你11|001|000。这是一个小测试：func TestSmokeTest(t *testing.T) {&nbsp; &nbsp; input := "11001000"&nbsp; &nbsp; s := partition(input, "|", 3)&nbsp; &nbsp; if s != "11|001|000" {&nbsp; &nbsp; &nbsp; &nbsp; t.Fail()&nbsp; &nbsp; }&nbsp; &nbsp; s = partition(input, "|", 2)&nbsp; &nbsp; if s != "11|00|10|00" {&nbsp; &nbsp; &nbsp; &nbsp; t.Fail()&nbsp; &nbsp; }&nbsp; &nbsp; input = "0111001000"&nbsp; &nbsp; s = partition(input, "|", 3)&nbsp; &nbsp; if s != "0|111|001|000" {&nbsp; &nbsp; &nbsp; &nbsp; t.Fail()&nbsp; &nbsp; }&nbsp; &nbsp; s = partition(input, "|", 2)&nbsp; &nbsp; if s != "01|11|00|10|00" {&nbsp; &nbsp; &nbsp; &nbsp; t.Fail()&nbsp; &nbsp; }}