华为上机测试题(大数相乘-java)

PS:这个不是自己写的,测试OK,供参考。

/**
* 大数相乘
*/

  1 public class BigData {
  2 
  3     public static void main(String[] args) {
  4         // TODO Auto-generated method stub
  5          StringBuffer sb = new StringBuffer();
  6 //        String str1 = "99999999999";
  7 //        String str2 = "99999999999";
  8          for (int i = 0; i < 2000; i++) {
  9          sb.append(9);
 10          }
 11          String str1 = sb.toString();
 12          String str2 = str1;
 13          BigData test = new BigData();
 14         String result = test.mutiStr(str1, str2);
 15         System.out.println(result);
 16     }
 17     
 18     public String mutiStr(String str1, String str2) {
 19         String result = null;// 存放结果字符串
 20 
 21         // 确定负号
 22         int flagStr1 = 0;
 23         if (str1.startsWith("-")) {
 24             flagStr1++;
 25         }
 26         int flagStr2 = 0;
 27         if (str2.startsWith("-")) {
 28             flagStr2++;
 29         }
 30         // *********************************************************************
 31 
 32         // 将数字放到整数数组,逆序,字符串尾部对齐
 33         int[] arrStr1 = new int[str1.length() - flagStr1];
 34         for (int i = 0; i < arrStr1.length; i++) {
 35             arrStr1[i] = str1.charAt(str1.length() - 1 - i) - '0';
 36         }
 37         int[] arrStr2 = new int[str2.length() - flagStr2];
 38         for (int i = 0; i < arrStr2.length; i++) {
 39             arrStr2[i] = str2.charAt(str2.length() - 1 - i) - '0';
 40         }
 41         // ***********************************************************************
 42 
 43         // 运算开始,arrStr2的每位去乘arrStr1的每位
 44         StringBuffer subResult = new StringBuffer();// 中间结果
 45         for (int i = 0; i < arrStr2.length; i++) {
 46             StringBuffer midResult = new StringBuffer();// 每位结果
 47             int subBenWei = 0;// 每位相乘时的本位上数字
 48             int subJinWei = 0;// 每位相乘时的进位上数字
 49             for (int k = 0; k < i; k++) {
 50                 midResult.append(0);// 先添加0,确保每位末尾对齐,便于加法运算
 51             }
 52             for (int j = 0; j < arrStr1.length; j++) {
 53                 subBenWei = (arrStr2[i] * arrStr1[j] + subJinWei) % 10;
 54                 subJinWei = (arrStr2[i] * arrStr1[j] + subJinWei) / 10;
 55                 midResult.append(subBenWei);
 56                 if ((j == arrStr1.length - 1) && subJinWei > 0) {
 57                     midResult.append(subJinWei);// 如果最前面位相乘有进位,结果添加进位数字
 58                 }
 59             }
 60             if (0 == i) {
 61                 subResult = midResult;// 第一位,子结果存放第一位结果
 62             } else {
 63                 // 各位对齐,确保中间结果和每位结果位数相同,便于相加运算
 64                 if (midResult.length() > subResult.length()) {
 65                     int diff = midResult.length() - subResult.length();
 66                     while (diff > 0) {
 67                         subResult.append(0);
 68                         diff--;
 69                     }
 70                 } else {
 71                     int diff = subResult.length() - midResult.length();
 72                     while (diff > 0) {
 73                         midResult.append(0);
 74                         diff--;
 75                     }
 76                 }
 77                 // *********************************************************
 78                 // 相加运算,结果放到subResult中
 79                 int benWei = 0;
 80                 int jinWei = 0;
 81                 for (int j = 0; j < subResult.length(); j++) {
 82 
 83                     benWei = ((subResult.charAt(j) - '0')
 84                             + (midResult.charAt(j) - '0') + jinWei) % 10;
 85                     jinWei = ((subResult.charAt(j) - '0')
 86                             + (midResult.charAt(j) - '0') + jinWei) / 10;
 87                     subResult.setCharAt(j, (char) (benWei + '0'));
 88                 }
 89                 // 如果最后相加任有进位,结果中添加进位
 90                 if (jinWei > 0) {
 91                     subResult.append(jinWei);
 92                 }
 93                 // ***********************************************************
 94             }
 95         }
 96         // 判断“-”的个数,没有或有两个则为结果为正,不考虑,如只有一个则结果要添加“-”
 97         if (1 == (flagStr1 + flagStr2)) {
 98             subResult.append('-');
 99         }
100         result = subResult.reverse().toString();// 最终子结果逆序,即为结果
101         return result;
102     }
103 }
原文地址:https://www.cnblogs.com/lr-ting/p/3998478.html