不使用解析值或数组的二进制到十进制转换器

Question

得到字符串索引超出范围，但我不明白为什么我已经通过了大约50次

import java.util.Scanner;
public class binary {
public static void main(String[] args) {
System.out.println("Enter the first binary number");
Scanner keyboard = new Scanner(System.in);
String num1 = keyboard.next();
//System.out.println("Enter the second binary number");
//String num2 = keyboard.next();

int total = 0;
for(int i = num1.length(); i>0;i--) {
if(num1.charAt(i) == 1) {
total += 2*i;
}

}
if(num1.charAt(3) == 1) {
total -= 1;
}
System.out.println(total);
}
}

Answer 1

这里有一个完整的解决方案，包括一组测试:

class binary {
private static int binaryToInt(String binary) {
int total = 0;
for (int i = 0 ; i < binary.length(); i++) {
total *= 2;
if (binary.charAt(i) == '1')
total += 1;
}
return total;
}
private static void test(String binary, int expected) {
int n = binaryToInt(binary);
String rightWrong = "right";
if (n != expected) {
rightWrong = String.format("WRONG! (should be %d)", expected);
System.out.printf("%s -> %d is %sn", binary, n, rightWrong);
}
public static void main(String[] args) {
test("0", 0);
test("1", 1);
test("10", 2);
test("100", 4);
test("111", 7);
test("0000111", 7);
test("1010101010", 682);
test("1111111111", 1023);
System.out.println("");
// test sanity check
System.out.println("This last test should fail (we are just testing the test method itself here)...");
test("1010101010", 0);
}
}

结果:

0 -> 0 is right
1 -> 1 is right
10 -> 2 is right
100 -> 4 is right
111 -> 7 is right
0000111 -> 7 is right
1010101010 -> 682 is right
1111111111 -> 1023 is right
This last test should fail (we are just testing the test method itself here)...
1010101010 -> 682 is WRONG! (should be 0)

代码中的一个重要问题尚未在注释或先前的回答中解决。注意这一行和你代码中的那一行:

if (binary.charAt(i) == '1')

您正在测试数值1，它永远不会是true，因为您从charAt()返回的是一个字符，而不是一个数字。

Answer 2

length()计算元素的个数，它们的索引从0开始。对于字符串"1111"最后一个字符应该在索引3，而不是4，所以是.length()-1。您需要将for语句更改为for(int i = num1.length()-1; i>=0;i--)(也注意条件更改)，或者将charAt语句更改为if(num1.charAt(i-1) == '1')。

同样，根据你想做的，我假设对于total += 2*i，你实际上需要像total += Math.pow(2, i-length())这样的东西，这取决于你决定先用i做什么。