Number systems have different bases. Decimals numbers with a base of ten are what most of us are familiar with. Computers, on the other hand, use only ones and zeros (binary). Let's try to write some code that converts binary numbers to decimals.

Specifically, we want to develop an algorithm that accepts a string containing ones and zeros and returns an integer.

We can convert the binary string by following these steps:

1. Start from the end of the string and process each character at a time. The position of each digit in the binary string corresponds to a decimal number in a sequence.
2. To generate this sequence, you start from one and multiply by two every time, so one, two, four, eight, and so on (see Conversion Sequence row of Table 1.1). More formally, the sequence is a geometric progression that starts at one and progresses in a common ratio of two.
3. We then apply the binary string as a mask on this sequence (see the Binary String (Mask) row of Table 1.1).
4. The result is a new sequence where the values are only kept if the corresponding position in the binary string has a value of one (see the Result row of Table 1.1).
5. After applying the mask, we just need to sum up the resulting numbers together.

In the preceding example (Table 1.1), resulting total is 22. This is our decimal number corresponding to the binary number 10110.

Snippet 1.1 shows us how we can do this. We use a single conversion variable instead of a sequence and initialize this variable to the value of one. We then use a loop to iterate over the length of the binary string starting from the end. While iterating, if the digit at our current position is one, we add the current conversion variable to the final result. We then simply double the current conversion variable and repeat. The code snippet is as follows:

public int convertToDecimal(String binary) {
  int conversion = 1;
  int result = 0;
  for (int i = 1; i <= binary.length(); i++) {
    if (binary.charAt(binary.length() - i) == '1')
      result += conversion;
    conversion *= 2;
  }
  return result;
}