在Java中,将int类型转换为二进制字符串的最佳方法(理想情况下最简单)是什么?
例如,如果要转换的int值为156,则其二进制字符串表示形式应为“10011100”。
Integer.toBinaryString(int i)
还有一个java.lang.Integer.toString(int i, int base)方法,如果您的代码将来可能处理二进制以外的其他进制,则此方法更适合。请注意,该方法仅为整数i提供无符号表示,如果i为负数,则在前面添加一个负号。它不会使用二进制补码。
public static string intToBinary(int n)
{
String s = "";
while (n > 0)
{
s = ( (n % 2 ) == 0 ? "0" : "1") +s;
n = n / 2;
}
return s;
}
另一种方法-通过使用java.lang.Integer,您可以获取第一个参数i
的字符串表示形式,该字符串表示形式由第二个参数指定的基数(八进制-8,十六进制-16,二进制-2)
来确定。
Integer.toString(i, radix)
例子_
private void getStrtingRadix() {
// TODO Auto-generated method stub
/* returns the string representation of the
unsigned integer in concern radix*/
System.out.println("Binary eqivalent of 100 = " + Integer.toString(100, 2));
System.out.println("Octal eqivalent of 100 = " + Integer.toString(100, 8));
System.out.println("Decimal eqivalent of 100 = " + Integer.toString(100, 10));
System.out.println("Hexadecimal eqivalent of 100 = " + Integer.toString(100, 16));
}
输出_
Binary eqivalent of 100 = 1100100
Octal eqivalent of 100 = 144
Decimal eqivalent of 100 = 100
Hexadecimal eqivalent of 100 = 64
public class Main {
public static String toBinary(int n, int l ) throws Exception {
double pow = Math.pow(2, l);
StringBuilder binary = new StringBuilder();
if ( pow < n ) {
throw new Exception("The length must be big from number ");
}
int shift = l- 1;
for (; shift >= 0 ; shift--) {
int bit = (n >> shift) & 1;
if (bit == 1) {
binary.append("1");
} else {
binary.append("0");
}
}
return binary.toString();
}
public static void main(String[] args) throws Exception {
System.out.println(" binary = " + toBinary(7, 4));
System.out.println(" binary = " + Integer.toString(7,2));
}
}
将整数转换为二进制:
import java.util.Scanner;
public class IntegerToBinary {
public static void main(String[] args) {
Scanner input = new Scanner( System.in );
System.out.println("Enter Integer: ");
String integerString =input.nextLine();
System.out.println("Binary Number: "+Integer.toBinaryString(Integer.parseInt(integerString)));
}
}
输出:
输入整数:
10
二进制数:1010
最简单的方法是检查数字是否为奇数。如果是,根据定义,它的最右边的二进制数字将是“ 1”(2 ^ 0)。之后,我们将数字向右位移,并使用递归检查相同的值。
@Test
public void shouldPrintBinary() {
StringBuilder sb = new StringBuilder();
convert(1234, sb);
}
private void convert(int n, StringBuilder sb) {
if (n > 0) {
sb.append(n % 2);
convert(n >> 1, sb);
} else {
System.out.println(sb.reverse().toString());
}
}
这是我几分钟前随便写的东西。希望能有所帮助!
public class Main {
public static void main(String[] args) {
ArrayList<Integer> powers = new ArrayList<Integer>();
ArrayList<Integer> binaryStore = new ArrayList<Integer>();
powers.add(128);
powers.add(64);
powers.add(32);
powers.add(16);
powers.add(8);
powers.add(4);
powers.add(2);
powers.add(1);
Scanner sc = new Scanner(System.in);
System.out.println("Welcome to Paden9000 binary converter. Please enter an integer you wish to convert: ");
int input = sc.nextInt();
int printableInput = input;
for (int i : powers) {
if (input < i) {
binaryStore.add(0);
} else {
input = input - i;
binaryStore.add(1);
}
}
String newString= binaryStore.toString();
String finalOutput = newString.replace("[", "")
.replace(" ", "")
.replace("]", "")
.replace(",", "");
System.out.println("Integer value: " + printableInput + "\nBinary value: " + finalOutput);
sc.close();
}
}
使用内置函数:
String binaryNum = Integer.toBinaryString(int num);
如果你不想使用内置的将整数转换为二进制的函数,你也可以这样做:
import java.util.*;
public class IntToBinary {
public static void main(String[] args) {
Scanner d = new Scanner(System.in);
int n;
n = d.nextInt();
StringBuilder sb = new StringBuilder();
while(n > 0){
int r = n%2;
sb.append(r);
n = n/2;
}
System.out.println(sb.reverse());
}
}
这是我的方法,它有一个固定字节数的小优点。
private void printByte(int value) {
String currentBinary = Integer.toBinaryString(256 + value);
System.out.println(currentBinary.substring(currentBinary.length() - 8));
}
public int binaryToInteger(String binary) {
char[] numbers = binary.toCharArray();
int result = 0;
for(int i=numbers.length - 1; i>=0; i--)
if(numbers[i]=='1')
result += Math.pow(2, (numbers.length-i - 1));
return result;
}
String.format("%8s", Integer.toBinaryString(i)).replace(' ', '0')
- undefined