劍指offer——大數相乘問題(BigInteger的大致實現思路)
阿新 • • 發佈:2019-01-08
乘法運算可以分拆為兩步:第一步,是將乘數與被乘數逐位相乘;第二步,將逐位相乘得到的結果,對應相加起來。
public class BigInt {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str1 = sc.next();
String str2 = sc.next();
int[] num1 = new int[str1.length()];
int [] num2 = new int[str2.length()];
//把字串轉換成int陣列
for (int i = 0; i < str1.length(); i++) {
num1[str1.length() - 1 - i] = str1.charAt(i) - '0';
}
for (int i = 0; i < str2.length(); i++) {
num2[str2.length() - 1 - i] = str2.charAt(i) - '0' ;
}
int[] result = multiply(num1, num2);
// 因為前面做了倒序操作,所以最高位在最後
for (int i = result.length - 1; i >= 0; i--) {
if(i==result.length - 1)
if(result[i]==0)
continue;
System.out.print(result[i]);
}
sc.close();
}
public static int[] multiply(int[] num1, int[] num2) {
int lengthOfNum1 = num1.length;
int lengthOfNum2 = num2.length;
//如果num1和num2的長度分別為n1,n2,那麼它們相乘的結果位數最大為n1+n2
int[] result = new int[lengthOfNum1 + lengthOfNum2];
//num[i]*num2[j]的結果存在result[i+j]上,最後再處理進位制問題
for (int i = 0; i < lengthOfNum1; i++) {
for (int j = 0; j < lengthOfNum2; j++) {
result[i + j] += num1[i] * num2[j];
}
}
// 處理進位制問題
for (int i = 0; i < lengthOfNum1 + lengthOfNum2 - 1; i++) {
if (result[i] >= 10) {
result[i + 1] += result[i] / 10;
result[i] = result[i] % 10;
}
}
return result;
}
}