package arithmetic.com.ty.binary;

public class LargeAdd {

    /**
     * 大整数求和
     * 
     * @param bigNumberA 大整数A
     * @param bigNumberB 大整数B
     */
    public static String bigNumberSum(String bigNumberA, String bigNumberB) {
        //1.把两个大整数用数组逆序存储，数组长度等于较大整数位数+1
        int maxLength = bigNumberA.length() > bigNumberB.length() ? bigNumberA.length() : bigNumberB.length();
        int[] arrayA = new int[maxLength + 1];
        /**
         * 1.大整数倒着存放在数组中，因为计算得先从个位数开始计算，所以为了方便，将个位数存在下标0处，然后正常循环数组就可以做加操作
         * 2.获取大整数的每位char，通过  char - ‘0‘ 再转换成int
         */
        for (int i = 0; i < bigNumberA.length(); i++) {
            arrayA[i] = bigNumberA.charAt(bigNumberA.length() - 1 - i) - ‘0‘;
        }
        int[] arrayB = new int[maxLength + 1];
        for (int i = 0; i < bigNumberB.length(); i++) {
            arrayB[i] = bigNumberB.charAt(bigNumberB.length() - 1 - i) - ‘0‘;
        }
        //2.构建result数组，数组长度等于较大整数位数+1 
        int[] result = new int[maxLength + 1];
        //3.遍历数组，按位相加
        for (int i = 0; i < result.length; i++) {
            int temp = result[i];
            temp += arrayA[i];
            temp += arrayB[i];
            //判断是否进位
            if (temp >= 10) {
                temp = temp - 10;
                result[i + 1] = 1;
            }
            result[i] = temp;
        }

        // 4.把result数组再次逆序并转成String
        StringBuilder sb = new StringBuilder();
        //是否找到大整数的最高有效位
        boolean findFirst = false;
        for (int i = result.length - 1; i >= 0; i--) {
            if (!findFirst) {
                if (result[i] == 0) {
                    continue;
                }
                findFirst = true;
            }
            sb.append(result[i]);
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        System.out.println(bigNumberSum("426709752318", "95481253129"));
    }
}