两个超大数的加法-遇到的一个笔试题

最近碰到的一个笔试题,应该有多种实现方式,我也只是那一个最常用的方法搞定的
  1 import java.math.BigInteger;
  2 import java.util.regex.Pattern;
  3 
  4 public class AddBigData {
  5 
  6     private static final int SINGLE_MAX = 9;
  7     private static final int ONE = 1;
  8 
  9     public static String addBigData(String fir, String sec) throws NumberFormatException{
 10         //check format is ok, first to check weather exist zero
 11         if("0".equals(fir)){
 12             return sec;
 13         }else if("0".equals(sec)){
 14             return fir;
 15         }else {
 16             //use regx to check number format
 17             Pattern dataPattern = Pattern.compile("^[1-9]\d*$");//start not with zero ,end with data
 18             boolean firCheck = dataPattern.matcher(fir).find();
 19             boolean secCheck = dataPattern.matcher(sec).find();
 20             
 21             if(firCheck && secCheck){
 22                 //begin to cal
 23                 return add(fir, sec);
 24             }else{
 25                 //throw exception
 26                 throw new NumberFormatException("输入的字符串格式错误!请输入正确的数字。");
 27             }
 28         }
 29     }
 30     
 31     public static String add(String fir, String sec) {
 32         StringBuilder res = new StringBuilder();
 33         
 34         char[] firs = fir.toCharArray();
 35         char[] secs = sec.toCharArray();
 36         //get min length to cal, get maxlen to store
 37         int minLen = Math.min(firs.length, secs.length);
 38         int maxLen = Math.max(firs.length, secs.length);
 39         
 40         int[] aiRes = new int[maxLen+1];//to store, be carefull the aiRes's length must be maxLen+1
 41         //init aiRes
 42         aiRes[0] = 0;
 43         char[] maxString = firs.length >= secs.length ? firs : secs;
 44         for(int i = 0; i < (maxLen - minLen); i++){
 45             aiRes[i+1] = Integer.parseInt(String.valueOf(maxString[i]));
 46         }
 47         
 48         int prompt = 0;//data to add in the next step
 49         //begin to cal
 50         for(int i = 0; i < minLen; i++){
 51             char cFir = firs[firs.length - 1 - i];
 52             char cSec = secs[secs.length - 1 - i];
 53             //change to number and add between (max-min) to max-1
 54             int iFir = Integer.parseInt(String.valueOf(cFir));
 55             int iSec = Integer.parseInt(String.valueOf(cSec));
 56             
 57             iFir += iSec;
 58             iFir += prompt;
 59             aiRes[aiRes.length - 1 - i] = iFir % 10;
 60             prompt = iFir / 10;
 61         }
 62         aiRes[maxLen - minLen] += prompt;
 63         //is needed to cal continue
 64         if(aiRes[maxLen - minLen] > SINGLE_MAX){
 65             aiRes[maxLen - minLen] %= 10;
 66             //cal the last of aiRes between 0 to max-min-1
 67             for(int i = maxLen - minLen - 1; i >= 0; i--){
 68                 aiRes[i] += ONE;
 69                 if (aiRes[i] > SINGLE_MAX) {
 70                     aiRes[i] %= 10;
 71                 }else{
 72                     break;
 73                 }
 74             }
 75         }
 76         
 77         if(aiRes[0] > 0){
 78             res.append(aiRes[0]);
 79         }
 80         for(int i = 1; i < aiRes.length; i++){
 81             res.append(aiRes[i]);
 82         }
 83         
 84         return res.toString();
 85     }
 86 
 87     /**
 88      * @param args
 89      */
 90     public static void main(String[] args) {
 91         String firstBigData = "19911999998888888888888888888888888888888888888888999999999999999999999999999999";
 92         String secBigData =    "1008000001111111111111111111111111111111111111111000000000000000000000000000001";
 93         System.out.println(addBigData(firstBigData, secBigData));
 94         BigInteger bi1 = new BigInteger(firstBigData);
 95         BigInteger bi2 = new BigInteger(secBigData);
 96         bi1 = bi1.add(bi2);
 97         System.out.println(bi1.toString());
 98         /*result: 20920000000000000000000000000000000000000000000000000000000000000000000000000000*/
 99     }
100 
101 }
Java For BigData
原文地址:https://www.cnblogs.com/craig-yilia/p/5165448.html