String to Integer (atoi)

/*
 * Problem : Parse an integer.
 * Input   : One line, may be empty.
 * Output  : The parsed number.
 * Approach: Parse by hand.
 * Time    : O(n)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static long solve(String[] lines) {
        // Step 1: Take the first line. An empty input has no line at all.
        String text = lines.length > 0 ? lines[0] : "";
        int index = 0;
        // Step 2: Skip the leading spaces.
        while (index < text.length() && text.charAt(index) == ' ') {
            index++;
        }
        // Step 3: Read an optional sign.
        long sign = 1;
        if (index < text.length() && (text.charAt(index) == '+' || text.charAt(index) == '-')) {
            sign = text.charAt(index) == '-' ? -1 : 1;
            index++;
        }
        // Step 4: Read digits. Stop early when the value is already too big, so a long cannot overflow.
        long value = 0;
        while (index < text.length() && Character.isDigit(text.charAt(index)) && value <= 3000000000L) {
            value = value * 10 + (text.charAt(index) - '0');
            index++;
        }
        // Step 5: Apply the sign and clamp to the 32-bit range.
        value *= sign;
        return Math.max(-2147483648L, Math.min(2147483647L, value));
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        java.util.List<String> all = new java.util.ArrayList<>();
        while (scanner.hasNextLine()) {
            all.add(scanner.nextLine());
        }
        String[] lines = all.toArray(new String[0]);
        System.out.println(solve(lines));
    }
}