题解 | #高精度整数加法#

import java.util.Scanner;
import java.lang.StringBuilder;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNext()) { // 注意 while 处理多个 case
            String a = in.next();
            String b = in.next();
            System.out.println(getSum(a, b));
        }
    }
    private static String getSum(String str1, String str2) {
        int lenDis = 0;
	  //用0给较短的字符串补位，让两个字符串长度相等，便于运算
        if (str1.length() > str2.length()) {
            lenDis = str1.length() - str2.length();
            for (int i = 0; i < lenDis; i++) {
                str2 = 0 + str2;
            }
        } else {
            lenDis = str2.length() - str1.length();
            for (int i = 0; i < lenDis; i++) {
                str1 = 0 + str1;
            }
        }
        StringBuilder sb = new StringBuilder();
        char [] arr1 = str1.toCharArray();
        char [] arr2 = str2.toCharArray();

        int localValue = 0;
        int jinwei = 0;
        for (int i = arr2.length - 1; i >= 0; i--) {
            //从最右边对两个字符串的最后一个字符依次相加，如果大于等于10，则进位为1，否则进位为0，并计算当前位的结果
            localValue = arr1[i]  + arr2[i] - '0' - '0' + jinwei;
            if (localValue >= 10) {
                jinwei = 1;
                localValue = localValue - 10;
            }else{
                jinwei = 0;
            }
            sb.append(localValue);
        }
	  //判断最高位相加是否进位，如果进位，那就在最高位加1
        if(jinwei>0){
            sb.append(1);
        }
        return sb.reverse().toString();
    }
}