题解 | #高精度整数加法#
高精度整数加法
https://www.nowcoder.com/practice/49e772ab08994a96980f9618892e55b6
import java.util.Scanner;
// 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();
if(a.length()<b.length()){
String temp = b;
b=a;
a=temp;
}
//记录较短数字的长度
int bLength = b.length()-1;
int jw = 0;
StringBuilder result = new StringBuilder ();
//按照长数字的长度从右到左进行遍历
for(int i=a.length()-1; i>=0; i--){
int sumSub = Integer.parseInt(a.substring(i,i+1));
//短长度数字是否遍历完
if(bLength>=0){
//没遍历完长数字最右一位+短数字最右一位+进位
sumSub = sumSub +Integer.parseInt(b.substring(bLength,bLength+1))+jw;
}else{
//短数字已遍历完,直接用长数字最有一位加上仅为
sumSub = sumSub+jw;
}
//所得数字大于10则进位为1,小于10进位为0
jw = sumSub>=10?1:0;
String sumString = sumSub+"";
//判断是否为最左一位
if(i>0){
result.append(sumString.substring(sumString.length()-1,sumString.length()));
}else{
result.append(new StringBuilder (sumString).reverse().toString());
}
bLength = bLength-1;
}
System.out.println(result.reverse().toString());
}
}
}

