趋势科技第一题
借用下群里老哥发的题图
import java.util.Scanner;
public class Adder {
// please don't modify any code below.
public static void main(String[] args) {
// Scanner sc = new Scanner(System.in);
// String num1 = sc.nextLine();
// String num2 = sc.nextLine();
String num1 = "199.999";
String num2 = "0.001";
String sum = add(num1, num2);
System.out.println(sum);
// System.out.println("2234567890223456789022345678902234567890.223456789022345678902234567890223456789");
}
private static String[] split(String str) {
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == '.') {
return new String[]{str.substring(0, i), str.substring(i + 1)};
}
}
return new String[]{str, ""};
}
private static String add(String num1, String num2) {
// 坑1 ：如何拿到正确的数
String[] number1 = split(num1);
String[] number2 = split(num2);
String Integer1 = number1[0];
String Integer2 = number2[0];
String dot1 = number1[1];
String dot2 = number2[1];
int jw = 0;
StringBuilder dotsb1 = new StringBuilder(dot1);
StringBuilder dotsb2 = new StringBuilder(dot2);
// 坑2：小数点尾数对不齐低的尾要补齐
while (dotsb1.length() != dotsb2.length()) {
if (dotsb1.length() < dotsb2.length()) {
dotsb1.append('0');
} else {
dotsb2.append('0');
}
}
dotsb1.reverse();
dotsb2.reverse();
StringBuilder integerRes = new StringBuilder();
int len2 = dotsb1.length();
boolean flag = true;
for (int i = 0; i < len2; i++) {
int tmp = jw;
if (i < dotsb1.length()) {
tmp = dotsb1.charAt(i) - '0' + tmp;
}
if (i < dotsb2.length()) {
tmp = dotsb2.charAt(i) - '0' + tmp;
}
if (tmp >= 10) {
jw = tmp / 10;
tmp %= 10;
} else {
jw = 0;
}
char c = (char) (tmp + '0');
// 坑3：刚开始加可能都是 0
if (flag && c == '0') {
continue;
} else if (c != '0') {
flag = false;
}
integerRes.append(c);
}
// 坑4：小数位可能全是 0
if (!flag) {
integerRes.append(".");
}
StringBuilder sb1 = new StringBuilder(Integer1);
sb1.reverse();
StringBuilder sb2 = new StringBuilder(Integer2);
sb2.reverse();
int len1 = sb1.length() > sb2.length() ? sb1.length() : sb2.length();
for (int i = 0; i < len1; i++) {
int tmp = jw;
if (i < sb1.length()) {
tmp = sb1.charAt(i) - '0' + tmp;
}
if (i < sb2.length()) {
tmp = sb2.charAt(i) - '0' + tmp;
}
if (tmp >= 10) {
jw = tmp / 10;
tmp %= 10;
} else {
jw = 0;
}
integerRes.append((char) (tmp + '0'));
}
// 坑5：最后一个进位
if (jw == 1) {
integerRes.append('1');
}
// 最后倒回去
return integerRes.reverse().toString();
}
}
#趋势科技##笔试题目#
