一次面试的经历，​设计一个加法，满足100位的数字相加？

public static String bigNumSum(String bigNumA,String bigNumB) {
    //1. String转为char数组
    //bigNumA, bigNumB 翻转两个字符串后，个位十位相加比较方便
    char[] bigNumAChars = new StringBuffer(bigNumA).reverse().toString().toCharArray();
    char[] bigNumBChars = new StringBuffer(bigNumB).reverse().toString().toCharArray();
    //2. 在长的数组长度上加一来存结果
    int resultLength;
    if (bigNumAChars.length > bigNumBChars.length) {
        resultLength = bigNumAChars.length;
    }else {
        resultLength = bigNumBChars.length;
    }
    int[] result = new int[resultLength + 1];
    //3.对位相加
    for (int i = 0; i < resultLength; i++) {
        // 如果当前的i超过了某个数组的长度，就用0代替高位了，和另一个字符数组中的数字相加
        int Aint = i < bigNumAChars.length ? (bigNumAChars[i] - '0') : 0;
        int Bint = i < bigNumBChars.length ? (bigNumBChars[i] - '0') : 0;
        int temp = result[i];
        temp += Aint;
        temp += Bint;
        result[i] = temp;
        //判断是否进位
        if (temp >= 10) {
            result[i + 1] = temp / 10;
            result[i] = temp % 10;
        }
    }
    //4. 存储最后的结果
    StringBuffer sb = new StringBuffer();
    //判断最高位是0还是1, 0无需保存
    if (result[result.length - 1] == 1) {
        sb.append(1);
    }
    for (int i = result.length-2; i >= 0; i--) {
        sb.append(result[i]);
    }
    return sb.toString();
}
public static void main(String[] args) {
    System.out.println(bigNumSum("8765876219","8882423"));
}