java练习之大数字求和

求两个不超过200位的非负整数的和

一、使用BigInteger类、BigDecimal类

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String [] args){
        Scanner red =new Scanner(System.in);
        BigInteger n5 = red.nextBigInteger();
        BigInteger n6 = red.nextBigInteger();
        String str = n5.add(n6).toString(); //大数求和
        System.out.println(str);
    }
}
运行结果：
输入：222222222222222222222222222222
输入：333333333333333333333333333333
555555555555555555555555555555

二、反转字符串、对齐字符串缺位补0、将两个正整数相加

public class AddTwoNumber {

 public static String add(String n1,String n2){

  String result=”“;

  //反转字符串
  String num1=new StringBuffer(n1).reverse().toString();
  String num2=new StringBuffer(n2).reverse().toString();

  int len1=num1.length();
  int len2=num2.length();
  int maxLen=len1>len2?len1:len2;
  //定义和(可能)
  int nSum[]=new int[maxLen+1];

  boolean nOverFlow=false;

  //对齐字符串
  if(len1for (int i = len1; i < len2; i++) {
    num1+=”0”;
   }
  }else if(len1>len2){
   for (int i = len2; i < len1; i++) {
    num2+=”0”;
   }
  }

  //两个数相加
  for (int i = 0; i < maxLen; i++) {
   //进位数从第二次开始算
   if (nOverFlow) {
    nSum[i]=Integer.parseInt(num1.charAt(i)+”“)+
      Integer.parseInt(num2.charAt(i)+”“)+1;
   }else{
    nSum[i]=Integer.parseInt(num1.charAt(i)+”“)+
      Integer.parseInt(num2.charAt(i)+”“);
   }
   //处理溢出位
   nOverFlow=handleSumOverTen(nSum,i);

  }

  //处理最高位  
  if(nOverFlow) {  
   nSum[maxLen] = 1;  
        }else {  
         nSum[maxLen] =0 ;  
        }

  for (int i = 0; i < nSum.length; i++) {
   result+=String.valueOf(nSum[i]);
  }
  String result1=new StringBuffer(result).reverse().toString();
  return result1;
 }

 private static boolean handleSumOverTen(int[] nSum, int i) {

  boolean flag = false;
  if(nSum[i] >= 10) {  
   nSum[i] = nSum[i] - 10;  
            flag = true;  
        }  
        else {  
            flag = false;  
        }  
        return flag;
 }



 public static void main(String[] args) {

  String num=add(“8888899999999888”, “88888888888888”);
  System.out.println(num);

 }

}

三、补齐字符串(使用StringBuffere中的insert方法在字符串索引为0的位置插入len个0)、对齐相加

public class BigNumSum2 {  
    public static void main(String[] args) {  
        int[] result = bigNumSum(“8888899999999888”, “88888888888888”);  
        for(int i=0; i < result.length; i++) {  
            System.out.print(result[i]);  
        }  
    }  

    public static int[] bigNumSum(String num1, String num2) {  

        String number1 = num1;  
        String number2 = num2; 

        int len1=number1.length();
        int len2=number2.length();
        int len=Math.abs(len1-len2);
        char insertNum[]=new char[len];
        for (int i = 0; i < insertNum.length; i++) {
   insertNum[i]=’0’;
  }
        String str1=”“;
        String str2=”“;
        //补齐两个字符串
        if (len1new StringBuffer(number1).insert(0, insertNum).toString();
         str2=number2;
  }else if(len1>len2){
   str1=number1;
         str2=new StringBuffer(number2).insert(0, insertNum).toString();
  }

        //字符串转换成字符数组
        char[] ch1 = str1.toCharArray();  
        char[] ch2 = str2.toCharArray();  
        int[] sum;  
        //为true时表示两数相加>=10 
        boolean flag = false;  

      //相加结果的长度为任一长度+1，因为最高位相加可能>10
        sum = new int[ch1.length+1];   
        //从个位开始相加  
        for(int i=ch1.length-1; i>=0; i–) {   
         //如果上一次相加和大于1，本次相加结果加1
            if(flag) {  
             //
                 sum[i+1] = (int)(ch1[i] - ‘0’) + (int)(ch2[i] - ‘0’) + 1;  
            }else {  
                    sum[i+1] = (int)(ch1[i] - ‘0’) + (int)(ch2[i] - ‘0’);  
            }  
            flag = handleSumOverTen(sum, i); //处理两数相加是否>10  
        }  

        handleTopDigit(flag, sum); //处理最高位  
        return sum;


    }  

    /* 
     * 处理两数相加是否>10 
     */  
    public static boolean handleSumOverTen(int[] sum, int i) {  
        boolean flag = false;  
        if(sum[i+1] >= 10) {  
            sum[i+1] = sum[i+1] - 10;  
            flag = true;  
        }  
        else {  
            flag = false;  
        }  
        return flag;  
    }  

    /* 
     * 处理最高位 
     */  
    public static void handleTopDigit(Boolean flag, int[] sum) {  
        if(flag) {  
            sum[0] = 1;  
        }else {  
            sum[0] = 0;  
        }  
    }  
}

四、四分法

public class BigNumSum {  
    public static void main(String[] args) {  
        int[] result = bigNumSum(“8888899999999888”, “88888888888888”);  
        for(int i=0; i < result.length; i++) {  
            System.out.print(result[i]);  
        }  
    }  

    public static int[] bigNumSum(String num1, String num2) {  

        String number1 = num1;  
        String number2 = num2; 
        //字符串转换成字符数组
        char[] ch1 = number1.toCharArray();  
        char[] ch2 = number2.toCharArray();  
        int[] sum;  
        //取位数之差
        int len = Math.abs(ch1.length - ch2.length);  
        //为true时表示两数相加>=10 
        boolean flag = false;  

      //如果两个数的长度相等  
      if(ch1.length == ch2.length) { 

        //相加结果的长度为任一长度+1，因为最高位相加可能>10
        sum = new int[ch1.length+1];   
        //从个位开始相加  
        for(int i=ch1.length-1; i>=0; i–) {   
         //如果上一次相加和大于1，本次相加结果加1
            if(flag) {  
             //
                 sum[i+1] = (int)(ch1[i] - ‘0’) + (int)(ch2[i] - ‘0’) + 1;  
            }else {  
                    sum[i+1] = (int)(ch1[i] - ‘0’) + (int)(ch2[i] - ‘0’);  
            }  
            flag = handleSumOverTen(sum, i, len); //处理两数相加是否>10  
        }  

        handleTopDigit(flag, sum); //处理最高位  
        return sum;  
      }  
        else if(ch1.length > ch2.length) { //如果数1的长度大于数2的长度  
            sum = new int[ch1.length+1]; //结果的长度为数1的长度+1  

            for(int i=ch2.length-1; i>=0; i–) {  
                if(flag) {  
                    sum[i+len+1] = (int)(ch1[i+len] - ‘0’) + (int)(ch2[i] - ‘0’) + 1;  
                }  
                else {  
                    sum[i+len+1] = (int)(ch1[i+len] - ‘0’) + (int)(ch2[i] - ‘0’);  
                }  

                flag = handleSumOverTen(sum, i, len);  
            }  

            for(int i=ch1.length-ch2.length-1; i>=0; i–) { //处理数1多出来的位数  
                if(flag) {  
                    sum[i+1] = (int)(ch1[i] - ‘0’) + 1;  
                }  
                else {  
                    sum[i+1] = (int)(ch1[i] - ‘0’);  
                }  
                flag = handleSumOverTen(sum, i, 0);  
            }  

            handleTopDigit(flag, sum);  
            return sum;  
        }  
        else {  
            sum = new int[ch2.length+1];  

            for(int i=ch1.length-1; i>=0; i–) {  
                if(flag) {  
                    sum[i+len+1] = (int)(ch1[i] - ‘0’) + (int)(ch2[i+len] - ‘0’) + 1;  
                }  
                else {  
                    sum[i+len+1] = (int)(ch1[i] - ‘0’) + (int)(ch2[i+len] - ‘0’);  
                }  

                flag = handleSumOverTen(sum, i, len);  
            }  

            for(int i=ch2.length-ch1.length-1; i>=0; i–) {  
                if(flag) {  
                    sum[i+1] = (int)(ch2[i] - ‘0’) + 1;  
                }  
                else {  
                    sum[i+1] = (int)(ch2[i] - ‘0’);  
                }  
                flag = handleSumOverTen(sum, i, 0);  
            }  

            handleTopDigit(flag, sum);  
            return sum;  
        }  
    }  

    /* 
     * 处理两数相加是否>10 
     */  
    public static boolean handleSumOverTen(int[] sum, int i, int len) {  
        boolean flag = false;  
        if(sum[i+len+1] >= 10) {  
            sum[i+len+1] = sum[i+len+1] - 10;  
            flag = true;  
        }  
        else {  
            flag = false;  
        }  
        return flag;  
    }  

    /*
     * 处理最高位 
     */  
    public static void handleTopDigit(Boolean flag, int[] sum) {  
        if(flag) {  
            sum[0] = 1;  
        }else {  
            sum[0] = 0;  
        }  
    }  
} 

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