java-两个大数相加

Posted 2020-08-14

题目要求：用字符串模拟两个大数相加。

一、使用BigInteger类、BigDecimal类

public static void main(String[] args) {

  String a="8888899999999888";
  String b="88888888888888";
  String str=new BigInteger(a).add(new BigInteger(b)).toString();
  System.out.println(str);
  
 }

 

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(len1<len2){
   for (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 (len1<len2) {
         
         str1=new 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、此方法与三中方法基本一致，不同之处是三中方法因为将两个字符串长度通过补0相等而不需要分步进行判断，但整体运行效率还是四分法高

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; 
        } 
    } 
}