/**
* Copyright (C), 2018-2020
* FileName: addStrings415
* Author: xjl
* Date: 2020/8/3 9:30
* Description: 给定两个字符串形式的非负整数 num1 和num2 ，计算它们的和。
*/
package String;
import java.util.Scanner;
public class addStrings415 {
public static String addStrings(String num1, String num2) {
//从尾部开始计算
int i = num1.length() - 1, j = num2.length() - 1, add = 0;
//存储结果
StringBuffer ans = new StringBuffer();
//遍历
while (i >= 0 || j >= 0 || add != 0) {
int x = i >= 0 ? num1.charAt(i) - '0' : 0;
int y = j >= 0 ? num2.charAt(j) - '0' : 0;
int result = x + y + add;
ans.append(result % 10);
add = result / 10;
i--;
j--;
}
// 计算完以后的答案需要翻转过来
ans.reverse();
return ans.toString();
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s1 = sc.nextLine();
String s2 = sc.nextLine();
String result = addStrings(s1, s2);
System.out.println(result);
}
}
其他的解法：利用的数组的方法
/**
* Copyright (C), 2018-2020
* FileName: MaxStringsum
* Author: xjl
* Date: 2020/7/27 12:57
* Description: 大数相加的和
*/
package Math;
import java.util.ArrayList;
import java.util.Scanner;
/**
* 18798458748987589654887584
* 875898748587968489578456984587
*/
public class MaxStringsum {
public static void main(String[] args) {
//数据的输入
Scanner sc = new Scanner(System.in);
String nums1 = sc.nextLine();
String nums2 = sc.nextLine();
//函数的调用
String result = testmax(nums1, nums2);
//结果的打印
System.out.println(result);
}
private static String testmax(String nums1, String nums2) {
nums1 = new StringBuffer(nums1).reverse().toString();
nums2 = new StringBuffer(nums2).reverse().toString();
//设置两个相同的数组存放数字
int[] array1 = new int[nums1.length()];
int[] array2 = new int[nums2.length()];
for (int i = 0; i < array1.length; i++) {
array1[i] = nums1.charAt(i) - '0';
}
for (int i = 0; i < array2.length; i++) {
array2[i] = nums2.charAt(i) - '0';
}
//存放结果
ArrayList<Integer> list = new ArrayList<>();
int i = 0;
int j = 0;
//进位
int current = 0;
int a = 0, b = 0;
while (i < array1.length || j < array2.length) {
if (i == array1.length) {
a = 0;
} else {
a = array1[i++];
}
if (j == array2.length) {
b = 0;
} else {
b = array2[j++];
}
int sum = a + b + current;
current = sum / 10;
list.add(sum % 10);
}
if (current == 1) {
list.add(current);
}
String res = "";
for (int V : list) {
res += String.valueOf(V);
}
return new StringBuffer(res).reverse().toString();
}
}
利用的BigInter函数来处理大数问题
/**
* Copyright (C), 2018-2020
* FileName: Bignumber
* Author: xjl
* Date: 2020/7/28 16:25
* Description: 大整数
*/
package Math;
import java.math.BigInteger;
import java.util.Scanner;
/**
* 相加：add(BigInteger val);
* 相减：subtract(BigInteger val);
* 相乘：multiply(BigInteger val);
* 相除：divide(BigInteger val);
* 最大公约数：gcd(BigInteger val);
* 取模：mod(BigInteger val);
* N次方：pow(int exponent);
*
*/
public class Bignumber {
public static void main(String[] args) {
//相加
Scanner sc = new Scanner(System.in);
String str1 = sc.nextLine();
String str2 = sc.nextLine();
BigInteger x1 = new BigInteger(str1);
BigInteger x2 = new BigInteger(str2);
//相加
BigInteger sum = x1.add(x2);
System.out.println(x1.add(x2));
//相乘
BigInteger and=x1.multiply(x2);
System.out.println(and);
}
}
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//初始进位为0
int pre = 0;
//操作数
ListNode mid = new ListNode(0);
//返回头节点
ListNode anws = mid ;
//当l1和l2都不为null时进入while循环
while(l1!=null&&l2!=null){
//操作数赋值
mid.val = (l1.val+l2.val+pre)%10;
//更新进位
pre = (l1.val+l2.val+pre)/10;
//更新头节点
l1 = l1.next;
l2=l2.next;
//头节点更新后判断是否为空
if(l1==null){
//如果l1头节点为空且进位为0，则操作数的next直接为l2剩下的
if(pre==0) {
mid.next = l2;
return anws;
}else {
//如果有进位，则递归调用addTwoNumbers方法
mid.next = addTwoNumbers(l2,new ListNode(pre));
return anws;
}
}
//同上
if(l2 == null){
if(pre==0) {
mid.next = l1;
return anws;
}else {
mid.next = addTwoNumbers(l1,new ListNode(pre));
return anws;
}
}
//l1 l2更新后都不为null，则设置操作数为0 进入下一次while循环
mid.next =new ListNode(0);
mid = mid.next;
}
//l1为null，直接不能进入上面while循环的情况下，直接返回l2
if(l1==null){
return l2;
}//同上
else if(l2 ==null){
return l1;
}
return anws;
}
}/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode pre = new ListNode(0);
ListNode cur = pre;
//进位符
int carry = 0;
while (l1 != null || l2 != null) {
//如果是等于空的时候要选择为0数进行加减
int x = l1 == null ? 0 : l1.val;
int y = l2 == null ? 0 : l2.val;
int sum=x + y + carry;
carry=sum/10;
int n =sum%10;
//添加新的节点
cur.next = new ListNode(n);
//并将向下移动
cur = cur.next;
//分别在两个中进行下一个的操作
if (l1 != null)
l1 = l1.next;
if (l2 != null)
l2 = l2.next;
}
if (carry == 1) {
cur.next = new ListNode(carry);
}
return pre.next;
}
}
