请设计一个算法能够完成两个用字符串存储的整数进行相加操作，对非法的输入则返回 error
数据范围：字符串长度满足 
[编程题]整数加法
import java.util.Scanner;
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 str1 = in.next();
String str2 = in.next();
System.out.println(caculate(str1,str2));
}
}
public static String caculate(String s1,String s2){
int i = s1.length() - 1,j = s2.length() - 1;
int carry = 0;
StringBuilder sb = new StringBuilder();
while(i >= 0 || j >= 0 || carry != 0){
if(i >= 0 && !Character.isDigit(s1.charAt(i)) || j >= 0 && !Character.isDigit(s2.charAt(j))) return "error";
int m = i < 0 ? 0 : s1.charAt(i) - '0';
int n = j < 0 ? 0 : s2.charAt(j) - '0';
int sum = m + n + carry;
sb.append(sum % 10);
carry = sum / 10;
i --;
j --;
}
return sb.reverse().toString();
}
}
发表于 2021-08-13 15:13:27
回复(0)
利用StringBuilder逆序按位算和就可以了
前排吐槽：用例给的是Error，然后题目中给的是error，提交错到这里太离谱了，也是我太不用心了
import java.util.*;
import java.math.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
int res = 0;
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String str;
while ((str = bf.readLine()) != null) {
String[] s = str.split(" ");
String s1 = new StringBuilder(s[0]).reverse().toString();
String s2 = new StringBuilder(s[1]).reverse().toString();
int index = 0;
boolean f = true;
boolean flag = false;
StringBuilder sb = new StringBuilder();
while (f && index < s1.length() && index < s2.length()) {
char ch1 = s1.charAt(index);
char ch2 = s2.charAt(index);
if (!Character.isDigit(ch1) || !Character.isDigit(ch2)) {
System.out.println("error");
f = false;
break;
}
int ten = ch1 - '0' + ch2 - '0';
if (flag) {
ten++;
flag = false;
}
flag = ten >= 10?true:false;
sb.append(ten % 10);
index++;
}
while (f && index < s1.length()) {
char ch1 = s1.charAt(index);
if (!Character.isDigit(ch1)) {
System.out.println("error");
f = false;
break;
}
int ten = ch1 - '0';
if (flag) {
ten++;
flag = false;
}
flag = ten >= 10?true:false;
sb.append(ten % 10);
index++;
}
while (f && index < s2.length()) {
char ch2 = s2.charAt(index);
if (!Character.isDigit(ch2)) {
System.out.println("error");
f = false;
break;
}
int ten = ch2 - '0';
if (flag) {
ten++;
flag = false;
}
flag = ten >= 10?true:false;
sb.append(ten % 10);
index++;
}
if (flag) {
sb.append(1);
}
if (f) {
System.out.println(sb.reverse().toString());
}
}
}
} 运行时间：7ms 占用内存：9236KB
编辑于 2020-12-24 10:29:56
回复(0)
用栈、链表、队列，都可以实现，不难理解
有一个小技巧是先让两个容器对齐，这样就不用处理特殊情况了，记得处理进位就可以了
import java.util.Scanner;
import java.util.Stack;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String str1=sc.next();
String str2=sc.next();
int len1=str1.length();
int len2=str2.length();
String result="";
Stack<Integer> stack1=new Stack<>();
Stack<Integer> stack2=new Stack<>();
Stack<Integer> stack3=new Stack<>();
//让两个栈对齐
if(len1>len2)
{
int count=len1-len2;
while(count-->0)
stack2.push(0);
}
else if(len1<len2)
{
int count=len2-len1;
while(count-->0)
stack1.push(0);
}
//分别入栈
for(int i=0;i<str1.length();i++)
{
char ch=str1.charAt(i);
if(ch<'0'||ch>'9')
{
System.out.println("error");
return;
}
stack1.push(ch-'0');
}
for(int i=0;i<str2.length();i++)
{
char ch=str2.charAt(i);
if(ch<'0'||ch>'9')
{
System.out.println("error");
return;
}
stack2.push(ch-'0');
}
//相加入新栈
int num1,num2,num3,add=0;
while( !stack1.isEmpty() && !stack2.isEmpty() )
{
num1=stack1.pop();
num2=stack2.pop();
num3=num1+num2+add;
add=num3/10;
num3%=10;
stack3.push(num3);
}
//处理最后的进位
if(add==1)
stack3.push(1);
while(!stack3.isEmpty())
result+=stack3.pop();
System.out.println(result);
}
}
发表于 2019-09-11 01:07:59
回复(0)
import java.util.Scanner;
public class Main { public static void main(String[] args) { String[] s = new Scanner(System.in).nextLine().split(" "); String a; String b; if(s[0].length()>s[1].length()){ a=s[0]; b=s[1]; }else{ a=s[1]; b=s[0]; } if((!a.matches("^[0-9]*$")) || (!b.matches("^[0-9]*$"))){ System.out.println("error");return; } int c=a.length()-b.length(); for (int i = 0; i <c ; i++) { b = "0" + b; } String result=""; int t=0,i=a.length(); while(i>0){ i--; int r=a.charAt(i)+b.charAt(i)-'0'-'0'+t; result=""+(r%10)+result; t=r/10; } System.out.println((t==0?"":t)+result); }
}
发表于 2018-11-06 09:42:05
回复(0)
//这上面的编译器无法识别包 是最无语的
import java.util.Scanner;
import java.math.BigInteger;
/**
* Created by Administrator on 2017/8/8.
*/
public class test {
public static String add(String a,String b){
for(int i=0;i<a.length();i++){//检查s1是否满足数字要求
if(!(a.charAt(i)>='0'&&a.charAt(i)<='9')){
return "error";
}
}
for(int i=0;i<b.length();i++){//检查s2是否满足数字
if(!(b.charAt(i)>='0'&&b.charAt(i)<='9')){
return "error";
}
}
BigInteger s1 = new BigInteger(a);
BigInteger s2 = new BigInteger(b);
s1=s1.add(s2);//a加上b，会有返回值的，返回的是BigInteger
return s1.toString();
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()){
String s=sc.nextLine();
String [] str=s.split(" ");
System.out.println(add(str[0],str[1]));
}
}
}
import java.util.Scanner;
import java.math.BigInteger;
/**
* Created by Administrator on 2017/8/8.
*/
public class test {
public static String add(String a,String b){
for(int i=0;i<a.length();i++){//检查s1是否满足数字要求
if(!(a.charAt(i)>='0'&&a.charAt(i)<='9')){
return "error";
}
}
for(int i=0;i<b.length();i++){//检查s2是否满足数字
if(!(b.charAt(i)>='0'&&b.charAt(i)<='9')){
return "error";
}
}
BigInteger s1 = new BigInteger(a);
BigInteger s2 = new BigInteger(b);
s1=s1.add(s2);//a加上b，会有返回值的，返回的是BigInteger
return s1.toString();
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()){
String s=sc.nextLine();
String [] str=s.split(" ");
System.out.println(add(str[0],str[1]));
}
}
}
发表于 2018-09-07 17:15:10
回复(0)
public class Test11 { public static void main(String args[]) { Scanner scanner = new Scanner(System.in); String c[] = scanner.nextLine().split(" "); char a[] = c[0].toCharArray(); char b[] = c[1].toCharArray(); int jingwei = 0; ArrayList<Integer> list = new ArrayList<>(); if (a.length>=b.length){ add(a, b, jingwei, list); } else{ add(b, a, jingwei, list); } for (int i=list.size()-1;i>=0;i--) System.out.print(list.get(i)); } private static boolean add(char[] a, char[] b, int jingwei, ArrayList<Integer> list) { for (int i=a.length-1,j=b.length-1;i>=0;i--,j--){ if (j>=0){ if (a[i]>'9'||a[i]<'0'||b[j]>'9'||b[j]<'0'){ System.out.println("error"); return true; } list.add((a[i]-'0'+b[j]-'0'+jingwei)%10); jingwei =(a[i]-'0'+b[j]-'0'+jingwei)/10; } else { if (a[i]>'9'||a[i]<'0'){ System.out.println("error"); return true; } list.add((a[i]-'0'+'0'-'0'+jingwei)%10); jingwei =(a[i]-'0'+'0'-'0'+jingwei)/10; } if (i==0&&jingwei!=0){ list.add(jingwei); } } return false; } }
发表于 2018-09-01 10:28:59
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
while(input.hasNext()){
char[] s1=input.next().trim().toCharArray();
char[] s2=input.next().trim().toCharArray();
Stack stack1=new Stack();
Stack stack2=new Stack();
for(int i=0;i<s1.length;i++){
int cur=0;
try{
cur=Integer.parseInt(String.valueOf(s1[i]));
}catch(Exception e){
System.out.println("error");
return;
}
stack1.push(cur);
}
for(int i=0;i<s2.length;i++){
int cur=0;
try{
cur=Integer.parseInt(String.valueOf(s2[i]));
}catch(Exception e){
System.out.println("error");
return;
}
stack2.push(cur);
}
Stack res=new Stack();
int a=0; //进位标志
while(!stack1.isEmpty()||!stack2.isEmpty()){
int a1=0;
if(!stack1.isEmpty()){
a1=stack1.pop();
}
int a2=0;
if(!stack2.isEmpty()){
a2=stack2.pop();
}
int cur=a1+a2+a;
if(cur>=10){
cur=cur%10;
a=1;
res.push(cur);
}else{
res.push(cur);
a=0;
}
}
//最后还要判断最高位是不是进位了
if(a==1){
res.push(a);
}
StringBuilder sb=new StringBuilder();
while(!res.isEmpty()){
sb.append(res.pop());
}
System.out.println(sb.toString());
}
}
}
发表于 2018-08-05 09:23:54
回复(0)
Java做的很累啊。尾部相加，进位保留往前加，加到没有进位为之。
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String a = scanner.next();
String b = scanner.next();
if (!legal(a) || !legal(b)) {
System.out.println("error");
}else {
System.out.println(add(a, b));
}
}
public static String add(String a,String b) {
int i = a.length() - 1;
int j = b.length() - 1;
int go = 0;
String res = "";
while (i >= 0 && j >= 0) {
int a1 = Integer.parseInt(a.substring(i, i + 1));
int b1 = Integer.parseInt(b.substring(j, j + 1));
int sum = a1 + b1 + go;
if (sum >= 10) {
go = 1;
sum = sum - 10;
}else {
go = 0;
}
res = sum + res;
i --;
j --;
}
String rem = i > j ? a.substring(0, i - j) : b.substring(0, j - i);
if (go == 0) {
res = rem + res;
}
int cur = rem.length() - 1;
while (cur >= 0) {
int num = Integer.parseInt(rem.substring(cur, cur + 1));
int sum = num + go;
if (sum >= 10) {
sum = sum - 10;
go = 1;
} else {
go = 0;
break;
}
res = String.valueOf(sum) + res;
cur --;
}
if (go == 1) {
res = "1" + res;
}
return res;
}
public static boolean legal(String s) {
for (int i = 0;i < s.length();i ++) {
if ('0' <= s.charAt(i) && s.charAt(i) <= '9') {
continue;
}else return false;
}
return true;
}
}
发表于 2018-06-14 00:21:26
回复(0)
运行时间：59ms
占用内存：9984k
一开始直接用的Integer，报case通过率80%，发现是输入的串很多个1超过了Integer范围，继续导入java.math.BigInteger搞定。
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static boolean is1(String str) {
for(int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if(c >= '0' && c <= '9') {
return true;
}
}
return false;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while(scan.hasNext()) {
String str = scan.nextLine();
String[] arr = str.split(" ");
if(is1(arr[0]) && is1(arr[1])) {
BigInteger a = new BigInteger(arr[0]);
BigInteger b = new BigInteger(arr[1]);
System.out.println(a.add(b));
}
else {
System.out.println("error");
}
}
}
}
看下面分析的，针对「abc123 123」这样的输入也会返回「error」的优化后的code：
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static boolean is1(String str) {
int count = 0;
for(int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if(c >= '0' && c <= '9') {
count++;
if(count == str.length()) {
count = 0;
return true;
}
}
}
return false;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while(scan.hasNext()) {
String str = scan.nextLine();
String[] arr = str.split(" ");
if(is1(arr[0]) && is1(arr[1])) {
BigInteger a = new BigInteger(arr[0]);
BigInteger b = new BigInteger(arr[1]);
System.out.println(a.add(b));
}
else {
System.out.println("error");
}
}
}
}
两段代码都通过了OJ。
编辑于 2018-04-13 13:40:30
回复(2)
import java.util.Scanner比短是比不过Python的，这辈子也比不过。
public static void main(String[] args) {
try {
Scanner input = new Scanner(System.in);
System.out.print(input.nextBigInteger().add(input.nextBigInteger()));
} catch (Exception e) {
System.out.print("error");
}
}
}
发表于 2018-04-10 13:48:37
回复(0)
import java.util.Stack;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
String s = scan.nextLine();
String[] ss = s.split(" ");
String result = getSumString(ss[0], ss[1]);
System.out.println(result);
}
}
public static String getSumString(String a ,String b) {
char[] s1 = a.toCharArray();
char[] s2 = b.toCharArray();
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>();
for(int i = 0 ; i< s1.length ; i++) {
if(new Integer(s1[i]) < 48 || new Integer(s1[i]) > 57) {
return "error";
}
stack1.push(new Integer(String.valueOf(s1[i])));
}
for(int i = 0 ; i< s2.length ; i++) {
if(new Integer(s2[i]) < 48 || new Integer(s2[i]) > 57) {
return "error";
}
stack2.push(new Integer(String.valueOf(s2[i])));
}
int k = 0;
Stack<Integer> stack3 = new Stack<Integer>();
while (!stack1.empty() || !stack2.empty()) {
Integer i = 0;
Integer j = 0;
if(!stack1.empty()) {
i = stack1.pop();
}
if(!stack2.empty()) {
j = stack2.pop();
}
if(i != null) {
k += i;
}
if(j != null) {
k += j;
}
stack3.push(k%10);
if(stack1.empty() && stack2.empty()) {
int s = k/10;
if (s > 0) {
stack3.push(s);
}
}
if(k > 9) {
k = 1;
} else {
k = 0;
}
}
StringBuffer stringBuffer = new StringBuffer();
while (!stack3.empty()) {
stringBuffer.append(stack3.pop());
}
return stringBuffer.toString();
}
}
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
String s = scan.nextLine();
String[] ss = s.split(" ");
String result = getSumString(ss[0], ss[1]);
System.out.println(result);
}
}
public static String getSumString(String a ,String b) {
char[] s1 = a.toCharArray();
char[] s2 = b.toCharArray();
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>();
for(int i = 0 ; i< s1.length ; i++) {
if(new Integer(s1[i]) < 48 || new Integer(s1[i]) > 57) {
return "error";
}
stack1.push(new Integer(String.valueOf(s1[i])));
}
for(int i = 0 ; i< s2.length ; i++) {
if(new Integer(s2[i]) < 48 || new Integer(s2[i]) > 57) {
return "error";
}
stack2.push(new Integer(String.valueOf(s2[i])));
}
int k = 0;
Stack<Integer> stack3 = new Stack<Integer>();
while (!stack1.empty() || !stack2.empty()) {
Integer i = 0;
Integer j = 0;
if(!stack1.empty()) {
i = stack1.pop();
}
if(!stack2.empty()) {
j = stack2.pop();
}
if(i != null) {
k += i;
}
if(j != null) {
k += j;
}
stack3.push(k%10);
if(stack1.empty() && stack2.empty()) {
int s = k/10;
if (s > 0) {
stack3.push(s);
}
}
if(k > 9) {
k = 1;
} else {
k = 0;
}
}
StringBuffer stringBuffer = new StringBuffer();
while (!stack3.empty()) {
stringBuffer.append(stack3.pop());
}
return stringBuffer.toString();
}
}
发表于 2018-03-23 11:19:43
回复(0)
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String string = null;
while ((string = br.readLine()) != null) {
//判断输入的字符串是否合法，如果合法则只包含数字和空格
boolean flag = false;
for (int i = 0; i < string.length(); i++) {
char ch = string.charAt(i);
if (ch != ' ' && (ch <= '0' || ch >= '9'))
flag = true;
}
if (flag) {
System.out.println("error");
continue;
}
//合法进行相加，因为数比较大，用int数组来存储数据
String[] strings = string.split(" ");
String s1, s2;
if (strings[0].length() > strings[1].length()) {
s1 = strings[0];
s2 = strings[1];
} else {
s1 = strings[1];
s2 = strings[0];
}
int[] result = new int[101];
int k = 0;
int sum = 0;
int add = 0;
int len1 = s1.length();
int len2 = s2.length();
while (k < len2) {
sum = (s1.charAt(len1-1-k)-'0') + (s2.charAt(len2 - 1- k)-'0') + add;
if (sum < 10) {
result[k] = sum;
add = 0;
} else {
result[k] = sum - 10;
add = 1;
}
k++;
}
while (k < len1) {
sum = (s1.charAt(len1 - 1 - k) - '0') + add;
if (sum < 10) {
result[k] = sum;
add = 0;
} else {
result[k] = sum - 10;
add = 1;
}
k++;
}
if (add == 1) {
result[k] = 1;
} else {
k--;
}
while (k >= 0) {
System.out.print(result[k]);
k--;
}
System.out.println();
}
}
}
发表于 2018-03-18 16:19:08
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()) {
String n1 = new StringBuffer(sc.next()).reverse().toString();
String n2 = new StringBuffer(sc.next()).reverse().toString();
String result = "";
int overflow = 0;
int i = 0;
while(i <= Math.max(n1.length(), n2.length())) {
if(i == Math.max(n1.length(), n2.length())) {
if(overflow != 0) result = overflow + result;
break;
}
int num1 = 0, num2 = 0;
if(i < n1.length()) num1 = n1.charAt(i) - '0';
if(i < n2.length()) num2 = n2.charAt(i) - '0';
if(num1 < 0 || num1 > 9 || num2 < 0 || num2 > 9) {
result = "error";
break;
}
int sum = num1 + num2 + overflow;
result = (sum % 10) + result;
overflow = sum / 10;
++i;
}
System.out.println(result);
}
}
}
发表于 2018-01-13 23:36:32
回复(0)
//这题让我知道有BigInteger的存在
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
try {
BigInteger num1 = in.nextBigInteger();
BigInteger num2 = in.nextBigInteger();
System.out.println(num2.add(num1));
} catch (Exception e) {
System.out.println("error");
}
}
}
----------------------------------------------------------------------
//自己实现大数的相加
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String regex = "\\d+";
while (in.hasNext()) {
StringBuilder str1 = new StringBuilder(in.next());
StringBuilder str2 = new StringBuilder(in.next());
if (str1.toString().matches(regex) && str2.toString().matches(regex)) {
//合法输入
processLen(str1, str2);
String res = add(str1, str2);
System.out.println(res);
} else {
System.out.println("error");
}
}
}
private static void processLen(StringBuilder str1, StringBuilder str2) {
int len1 = str1.length(), len2 = str2.length();
if (len1 > len2)
for (int i = 0; i < len1 - len2; i++) str2.insert(0, "0");
else
for (int i = 0; i < len2 - len1; i++) str1.insert(0, "0");
}
private static String add(StringBuilder str1, StringBuilder str2) {
StringBuilder res = new StringBuilder();
int reminder = 0;
int sum = 0;
for (int i = str1.length() - 1; i >= 0; i--) {
sum = str1.charAt(i) + str2.charAt(i) - 2 * '0' + reminder;
reminder = sum / 10;
if (i > 0)
res.append(sum % 10);
else {
res.append(sum % 10);
if (reminder > 0) res.append(1);
}
}
return res.reverse().toString();
}
}
------------------------------------------------------------------------
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String regex = "[+-]?\\d+";
while (in.hasNext()) {
String str1 = in.next();
String str2 = in.next();
if (str1.matches(regex) && str2.matches(regex)) {
BigInteger num1 = new BigInteger(str1);
BigInteger num2 = new BigInteger(str2);
System.out.println(num2.add(num1));
} else {
System.out.println("error");
}
}
}
}
编辑于 2017-12-26 13:55:37
回复(0)
import java.util.Scanner; public class Main { public static void main(String args[]) { Scanner scanner = new Scanner(System.in); while (scanner.hasNext()) { String[] strings = scanner.nextLine().split(" "); String string1 = strings[0]; String string2 = strings[1]; String str1; String str2; int length1 = string1.length(); int length2 = string2.length(); int a; int b; int c; int temp = 0; StringBuffer sb = new StringBuffer(); StringBuffer sb2 = new StringBuffer(); int x; int y; for (x = 0; x < length1; x++) { if (string1.charAt(x) < 48 || string1.charAt(x) > 57) { break; } } for (y = 0; y < length2; y++) { if (string2.charAt(y) < 48 || string2.charAt(y) > 57) { break; } } if (x == length1 && y == length2) { if (length1 == length2) { for (int i = length1 - 1; i >= 0; i--) { a = Integer.parseInt(String.valueOf(string1.charAt(i))); b = Integer.parseInt(String.valueOf(string2.charAt(i))); c = (a + b + temp) % 10; temp = (a + b + temp) / 10; sb.append(c); } if (temp != 0) { sb.append(temp); } } else { if (length1 > length2) { str1 = string1; str2 = string2; } else { int tem = length1; length1 = length2; length2 = tem; str1 = string2; str2 = string1; } int i = length1 - 1; int j = length2 - 1; for (; j >= 0; ) { a = Integer.parseInt(String.valueOf(str1.charAt(i))); b = Integer.parseInt(String.valueOf(str2.charAt(j))); c = (a + b + temp) % 10; temp = (a + b + temp) / 10; sb.append(c); if (j == 0) { if (i >= 2) { sb.append(Integer.parseInt(String.valueOf(str1.charAt(i - 1))) + temp); while (i >= 0) { sb.append(Integer.parseInt(String.valueOf(str1.charAt(i)))); i = i - 1; } break; } else { sb.append(Integer.parseInt(String.valueOf(str1.charAt(i - 1))) + temp); break; } } else { j = j - 1; i = i - 1; } } } for (int o = sb.length() - 1; o >= 0; o--) { sb2.append(sb.charAt(o)); } System.out.println(sb2.toString()); } else { System.out.println("error"); } } } }
这题主要用的想法就是小学数学的加法，超过10往前进位，我分了情况就是两个数字长度一样和不一样两种情况，还要注意非数字的情况
发表于 2017-11-30 20:27:07
回复(0)
public class Main{
public static String add(String a, String b){
int len1 = a.length();
int len2 = b.length();
//如果a的长度小于b，交换
if (len1 < len2){
String temp = a;
a = b;
b = temp;
}
len1 = a.length()-1;
len2 = b.length()-1;
StringBuilder sb = new StringBuilder();
//carry记录进位，sum记录每一位相加的和（取模后）
int carry = 0;
int sum = 0;
int x;int y;
while (len1 >=0){
x = a.charAt(len1)-'0';
if (len2 >=0){
y = b.charAt(len2) - '0';
}else {
y = 0;
}
//判断输入是否合法
if (x > 9 || x<0 || y<0 || y>9){
return "error";
}
sum = (x + y + carry)%10;
carry = (x + y + carry)/10;
len1--;len2--;
sb.append(sum);
}
if (carry > 0){
sb.append(carry);
}
return sb.reverse().toString();
}
public static void main(String[] args){
Scanner in = new Scanner(System.in);
String s[] = in.nextLine().split(" ");
if (s.length<2) {
System.out.println("error");
return;
}
System.out.println(add(s[0],s[1]));
}
}
发表于 2017-11-17 15:25:27
回复(0)
import java.util.Scanner;
//try-catch捕捉转换异常
public class IntegerPlus {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
StringBuilder sb = new StringBuilder();
while (sc.hasNext()) {
String s1 = sc.next();
String s2 = sc.next();
int temp1 = Math.max(s1.length(), s2.length());
int temp2 = Math.min(s1.length(), s2.length());
int[] jiashu = new int[temp1];// 加数集合
int[] beijiashu = new int[temp1];// 被加数集合
int[] jinwei = new int[temp1];// 进位数组集合
int[] result = new int[temp1];// 结果集合
try {// 数据转化出错，抛异常
if (s1.length() > s2.length()) {
for (int i = 0; i < temp1; i++)
jiashu[i] = Integer.valueOf(String.valueOf(s1.charAt(temp1 - i - 1)));
for (int j = 0; j < temp2; j++)
beijiashu[j] = Integer.valueOf(String.valueOf(s2.charAt(temp2 - j - 1)));// 补0
result[0] = (jiashu[0] + beijiashu[0]) % 10;
jinwei[0] = (jiashu[0] + beijiashu[0]) / 10;
sb.insert(0, result[0]);
for (int k = 1; k < temp1; k++) {
result[k] = (jiashu[k] + beijiashu[k] + jinwei[k - 1]) % 10;
jinwei[k] = (jiashu[k] + beijiashu[k] + jinwei[k - 1]) / 10;
sb.insert(0, result[k]);// 结果集合
}
if (jinwei[temp1 - 1] > 0)
sb.insert(0, 1);
} else {
for (int i = 0; i < temp1; i++)
jiashu[i] = Integer.valueOf(String.valueOf(s2.charAt(temp1 - i - 1)));
for (int j = 0; j < temp2; j++)
beijiashu[j] = Integer.valueOf(String.valueOf(s1.charAt(temp2 - j - 1)));// 补0
result[0] = (jiashu[0] + beijiashu[0]) % 10;
jinwei[0] = (jiashu[0] + beijiashu[0]) / 10;
sb.insert(0, result[0]);
for (int k = 1; k < temp1; k++) {
result[k] = (jiashu[k] + beijiashu[k] + jinwei[k - 1]) % 10;
jinwei[k] = (jiashu[k] + beijiashu[k] + jinwei[k - 1]) / 10;
sb.insert(0, result[k]);// 结果集合
}
if (jinwei[temp1 - 1] > 0)
sb.insert(0, 1);
}
System.out.println(sb.toString());
sb.setLength(0);
} catch (Exception e) {
System.out.println("Error");
sb.setLength(0);
}
}
}
}
发表于 2017-10-22 12:28:20
回复(0)
需要考虑几种情况：
1.非法字符串，这里不能直接使用try-catch捕获字符串转整形异常，因为字符串长度1-100大数字，数值保存肯定是不够的；
2.两个字符串的长度比较，然后从右往左进行同位两字符转整形相加，也就是把公有的位置数进行求和，然后在超出的那块长度与carry进位进行求和，最终还需要加上carry（我这里是用string保存结果，所以我加上carry）
3.在结果中使用StringBuffer对象的reverse反转过来，因为结果是反序保存的，考虑高位有可能有0的情况，需要使用正则:replace("^[0]*","")替换掉:
最后，代码有点长，才疏学浅，使用暴力了，不过浅显易懂，大神就跳过吧！轻点
import java.util.Scanner;
public class StringUtil {
//整数加法
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while(in.hasNext()){
String a = in.next();
String b = in.next();
int len = a.length();
int more = 0;
if(a.length() < b.length()){
len = b.length();
more = 1;
}
boolean ok = true;
//非法字符串判断
for(int i=0; i<len; i++){
if(i<a.length() && !(a.charAt(i)>='0' && a.charAt(i)<='9')){
System.out.println("error");
ok = false;
break;
}
if(i<b.length() && !(b.charAt(i)>='0' && b.charAt(i)<='9')){
System.out.println("error");
ok = false;
break;
}
}
if(ok == false)
continue;
//字符串后半部分相加:a更长
int carry = 0;
String res = "";
if(more == 0){
for(int i=len-1,j=b.length()-1; j>=0; j--){
int sum = Integer.valueOf(a.charAt(i--)+"") + Integer.valueOf(b.charAt(j)+"") + carry;
// System.out.println("sum="+sum);
carry = sum/10;
res += sum%10;
// System.out.println("res="+res);
// System.out.println("carry="+carry);
}
for(int i=len-b.length()-1; i>=0; i--){
int sum = Integer.valueOf(a.charAt(i)+"") + carry;
carry = sum/10;
res += sum%10;
// System.out.println("res="+res);
}
res += carry;
System.out.println(new StringBuffer(res).reverse().toString().replaceAll("^[0]*",""));
}
//字符串后半部分相加:b更长
else{
for(int i=len-1,j=a.length()-1; j>=0; j--){
int sum = Integer.valueOf(b.charAt(i--)+"") + Integer.valueOf(a.charAt(j)+"") + carry;
// System.out.println("sum="+sum);
carry = sum/10;
res += sum%10;
// System.out.println("res="+res);
// System.out.println("carry="+carry);
}
for(int i=len-a.length()-1; i>=0; i--){
int sum = Integer.valueOf(b.charAt(i)+"") + carry;
carry = sum/10;
res += sum%10;
// System.out.println("res="+res);
}
res += carry;
System.out.println(new StringBuffer(res).reverse().toString().replaceAll("^[0]*",""));
}
}
}
}
编辑于 2017-09-07 19:37:02
回复(0)
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