题解 | #高精度整数加法#
高精度整数加法
https://www.nowcoder.com/practice/49e772ab08994a96980f9618892e55b6
题解：使用字符串模拟加法运算，从尾到头进行运算，并将结果保存在StringBuilder中，运算完之后需要反转。难点在于对进位的处理。
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
Scanner scanner = new Scanner(br);
String str1 = scanner.nextLine();
String str2 = scanner.nextLine();
String result = getHighPrecisionIntAddition(str1,str2);
System.out.println(result);
}
private static String getHighPrecisionIntAddition(String str1, String str2) {
if (str1 == null || str1.length() == 0) return str2;
if (str2 == null || str2.length() == 0) return str1;
// 从尾到头计算低位的和
int i = str1.length()-1,j = str2.length()-1;
int carry = 0,sum;
StringBuilder builder = new StringBuilder();
while (i >= 0 || j >= 0){
if (i >= 0 && j >= 0){
sum = Integer.valueOf(str1.substring(i, i + 1)) + Integer.valueOf(str2.substring(j, j + 1));
i--;
j--;
}else if (i >= 0 && j < 0){
sum = Integer.valueOf(str1.substring(i, i + 1));
i--;
}else{
sum = Integer.valueOf(str2.substring(j, j + 1));
j--;
}
// 保存计算结果
builder.append((sum + carry) % 10);
// 处理进位
if ((sum + carry) >= 10){
carry = 1;
}else
carry = 0;
}
// 最终进位
if (carry == 1)
builder.append(carry);
// 结果需要反转
return builder.reverse().toString();
}
}
