【华为面试题】123456789*987654321用java实现，谢谢

要啥大整数:
直接搞:int a=123456789;
int b=987654321;
System.out.println((long)a*b);

1.java.math.biginteger
BigInteger b1 = new BigInteger("123456789");
BigInteger b2 = new BigInteger("987654321");
System.out.println(b1.multiply(b2).toString());
2.利用大数组或string都行，下面给出一个string的实现
ac poj 1001的片断 public String add(String s1,String s2) throws Exception{
if(s1 == null) throw new Exception("输入错误");
if(s2 == null) throw new Exception("输入错误");
String regex = "[\\d]+$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(s1);
if(!matcher.find()) throw new Exception("输入错误");
matcher = pattern.matcher(s2);
if(!matcher.find()) throw new Exception("输入错误");
String result1 = "";
int minLen =0;
int maxLen = 0;
String longString = "";
String shortString = "";
if(s1.length() > s2.length()) {
maxLen = s1.length();
minLen = s2.length();
longString = s1;
shortString = s2;
}else {
maxLen = s2.length();
minLen = s1.length();
longString = s2;
shortString = s1;
}
int carry = 0;
StringBuffer result = new StringBuffer();
int tempResult = 0;
for(int i = maxLen-1,j=minLen-1;j>-1;i--,j--) {
tempResult = Integer.parseInt(longString.substring(i, i+1)) + Integer.parseInt(shortString.substring(j, j+1)) + carry;
result.append(tempResult%10);
carry = (int)(tempResult/10);
}
for(int i=maxLen-minLen-1;i>-1;i--) {
tempResult = Integer.parseInt(longString.substring(i, i+1)) + carry;
result.append(tempResult%10);
carry = (int)(tempResult/10);
}
if(carry>0) {
result.append(carry);
}
return result.reverse().toString();
}
private String multi(String s1,String s2) throws Exception {
if(s1 == null) throw new Exception("输入错误");
if(s2 == null) throw new Exception("输入错误");
if(s2.length()!=1) throw new Exception("输入错误");
String regex = "[\\d]+$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(s1);
if(!matcher.find()) throw new Exception("输入错误");
matcher = pattern.matcher(s2);
if(!matcher.find()) throw new Exception("输入错误");
String result1 = "";
int tempResult = 0;
int carry = 0;
StringBuffer result = new StringBuffer();
int multi = Integer.parseInt(s2);
for(int i=s1.length()-1;i>-1;i--) {
tempResult = Integer.parseInt(s1.substring(i, i+1)) * multi + carry;
result.append(tempResult%10);
carry = (int)(tempResult/10);
}
if(carry>0) {
result.append(carry);
}
return result.reverse().toString();
}
public String multiple(String s1,String s2) throws Exception {
if(s1 == null) throw new Exception("输入错误");
if(s2 == null) throw new Exception("输入错误");
String regex = "[\\d]+$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(s1);
if(!matcher.find()) throw new Exception("输入错误");
matcher = pattern.matcher(s2);
if(!matcher.find()) throw new Exception("输入错误");
String result = "0";
String tempResult = "";
StringBuffer tempBuffer;
for(int i=0;i<s2.length();i++) {
tempResult = this.multi(s1, s2.substring(i, i+1));
tempBuffer = new StringBuffer(tempResult);
for(int j=i;j<s2.length()-1;j++) {
tempBuffer.append("0");
}
result = this.add(tempBuffer.toString(), result);
}
return result;
}

import java.math.BigInteger;
/**
*
* @author wubaochuan
*/
public class Main {
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        BigInteger a = new BigInteger("123456789");
        BigInteger b = new BigInteger("987654321");
        System.out.println(a.multiply(b));
        // TODO code application logic here
    }}

那究竟有啥要求, 有啥限制, 说清楚啊, 难道不能用long型, 哪个语言都支持这个吧

直接整说明你对long的取值范围很敏感。
他的结果是：121932631112635269
Long最大值：9223372036854775807
没错，是可以的。只有 2G*2G*2 以上的结果才会溢出。要是我出题，又想考算法的话，我就会附加一句（不能使用long数据类型）
要是我出题，是想考你们对数字范围的概念的话，我就像帖子标题这样讲。拜托，1234567890 还不到2G哩，987654321*987654321都不会溢出哦。我看直接整才是考官想要的吧

估计是要这样的吧  求的是 (123456789123456*98765432134567899)class big
{
public static void main(String[] args)
{
   int[] n1={1,2,3,4,5,6,7,8,9,1,2,3,4,5,6};
   int[] n2={9,8,7,6,5,4,3,2,1,3,4,5,6,7,8,9,9};
   int[] result=multiply(n1,n2);
   for(int i = 0; i < result.length; i++)
   {
      System.out.print(result[i]);
   }
  }
  private static int[] multiply(int[] numbers1, int[] numbers2)
  {
        int[] result = new int[numbers1.length + numbers2.length];
        for (int i = 0; i < numbers1.length; i++) {
            for (int j = 0; j < numbers2.length; j++) {
                int tenth = i + j;
                int cellResult = numbers1[i] * numbers2[j];
                put(result, tenth + 1, cellResult % 10);
                put(result, tenth, cellResult / 10);
            }
        }
        return result;
    }
     private static void put(int[] result, int index, int number) {
        result[index] += number;
        carryFrom(result, index);
    }
    private static void carryFrom(int[] result, int index) {
        if (index < 0) {
            return;
        }

        if (result[index] >= 10) {
            result[index - 1]++;
            result[index] = (int)(result[index] - 10);
            carryFrom(result, index - 1);
        }
    }   }

数组。。用String不行么。还算法这题目出的真纠结。。

那估计没啥好选的了，就是考大数相乘的算法了。输入是 string*string，输出还是string,里面用byte[], 算法还是标准乘法。

哈哈哈，被他欺骗啦，int很牛的,别小看它啦!!!

//用String好了
    public static String cm(String o, String t) {
        StringBuffer o1 = new StringBuffer(o).reverse();
        StringBuffer t1 = new StringBuffer(t).reverse();
        StringBuilder jg = new StringBuilder("0");
        int sum = 0;//当前位
        for (int i = 0, j; i < o1.length(); i++) {
            for (j = 0; j < t1.length(); j++) {
                sum += (o1.charAt(i) - '0') * (t1.charAt(j) - '0');
                if (jg.length() - 1 < i + j) {
                    jg.append('0');
                }
                sum += jg.charAt(i + j) - '0';
                jg.replace(i + j, i + j + 1, "" + (char) (sum % 10 + '0'));
                sum /= 10;
            }
            while (sum > 0) {
                if (jg.length() - 1 < i + j) {
                    jg.append('0');
                }
                sum += jg.charAt(i + j) - '0';
                jg.replace(i + j, i + j + 1, "" + (char) (sum % 10 + '0'));
                sum /= 10;
                j++;
            }
        }
        return jg.reverse().toString();
    }

调试易

【华为面试题】123456789*987654321用java实现，谢谢

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