乘法运算可以分拆为两步：第一步，是将乘数与被乘数逐位相乘；第二步，将逐位相乘得到的结果，对应相加起来。这有点类似小学数学中，计算乘法时通常采用的“竖式运算”。

package binary.search;

import java.util.Scanner;

public class Demo2 {
	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()];
		//把字符串转换成int数组，需要注意：为了操作方便，要进行倒序操作
		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);
		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的长度分别为n1，n2，那么它们相乘的结果位数为n1+n2-1到n1+n2
		int[] result = new int[lengthOfNum1 + lengthOfNum2];
		//num[i]*num2[j]的结果存在result[i+j]上，最后再处理进制问题
		for (int i = 0; i < lengthOfNum1; i++) {
			for (int j = 0; j < lengthOfNum2; j++) {
				result[i + j] += num1[i] * num2[j];
			}
		}
		// 处理进制问题
		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;
	}
}