实现有符号整数的二进制表示法。或者说，实现java.lang.Integer.toBinaryString()方法。
要想实现有符号整数的二进制表示法，我们首先需要知道有符号整数在计算机中是怎么存储的。
计算机中存储有符号整数，使用的是补码（two’s complement）。正数的补码同原码（其二进制表示）相同。负数的补码是其绝对值的原码按位取反（反码，one’s complement）后再加一。因此，在补码中只有一个0，也就是00000000000000000000000000000000。而10000000000000000000000000000000是最小的负数，在Java中也就是Integer.MIN_VALUE。
同时，这也带来一个我们在实现toBinaryString()函数时需要注意的问题，因为Java中Integer.MAX_VALUE，也就是01111111111111111111111111111111，的绝对值比Integer.MIN_VALUE小1。所以如果我们先求Integer.MIN_VALUE绝对值再求其二进制原码表示的话就会产生溢出。因此需要先将输入转化为Long才能避免这个问题。
代码实现如下：
/** * Implement java.lang.Integer.toBinaryString. */ public class ConvertBinary { public ConvertBinary() { super(); } /** * @param num The number to be converted to binary string * @return String representation of the 2's complement of the number. */ public String toBinaryString(int num) { Long new_num = Long.valueOf(num); // Case to Long to deal with Integer.MIN_VALUE if (num == 0) { return "0"; } else if (num > 0) { // the 2's complement of a positive number is the binary representation // of the number return toBaseTwo(new_num, false); } else { // num < 0 // the 2's complement of a negative number is the 1's complement of that number plus 1. // the 1's complement of a negative number can be obtained by reverting all the digits // of the base-two representation of it's absolute value. new_num *= -1; String result = toBaseTwo(new_num, true); result = revertDigit(result); result = addOne(result); return result; } } /** * @param num The number to be converted to base 2 representation * @param flag Boolean flag to indicates if 0s need to be complemented * to make the base 2 representation 32 digits long. This is needed for negative original inputs. * @return String representation of a base 2 representation. */ private String toBaseTwo(Long num, boolean flag) { StringBuilder sb = new StringBuilder(); while (num > 0) { long curr = num % 2; num = num / 2; sb.append(curr); } if (flag) { while (sb.length() < 32) { // add extra 0s to the binary string to make it 32 bits long sb.append('0'); } } return sb.reverse().toString(); } private String revertDigit(String num) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < num.length(); i++) { sb.append(num.charAt(i) == '0' ? '1' : '0'); } return sb.toString(); } private String addOne(String num) { StringBuilder sb = new StringBuilder(); int carryOver = 1; int i = num.length() - 1; for (; i >= 0; i--) { int curr = num.charAt(i) - '0'; curr += carryOver; if (curr == 2) { carryOver = 1; curr = 0; } else { carryOver = 0; } sb.append(curr); } return sb.reverse().toString(); } }