方法1 ：用兩個字串儲存輸入的大整數，然後用第二個字串的每一位去乘第一個字串的數字值，最後將每次的結果錯位相加即可。時間複雜度高O（n^2）

方法2：將兩個大整數X,Y每次分割成兩半，第一個分割成AB，第二個分割成CD。所以最後結果XY=（A*10^n/2 +B）（C*10^m/2+D）;進行分解可得

XY=AC*10^(n+m)/2+AD*10^n/2 +BC*10^m/2 +BD; 注意n/2,m/2為字串後半度的位數。分治法的時間複雜度為O(nlogn);

程式碼如下：兩個方法 放在同名的過載函式中。

package com.test;

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int n, m;
		String str[] = in.nextLine().split(" ");
		String result = multiBigInt(str[0], str[1]);

		// 除去首位的0
		char[] temp = result.toCharArray();
		int i = 0;
		for (; i < temp.length; i++) {
			if (temp[i] == '0')
				continue;
			else
				break;
		}
		System.out.println(result.substring(i, result.length()));

		String result2 = "";
		result2 = multiBigInt(str[0].toCharArray(), str[1].toCharArray());

		temp = result2.toCharArray();
		i = 0;
		for (; i < temp.length; i++) {
			if (temp[i] == '0')
				continue;
			else
				break;
		}
		System.out.println(result2.substring(i, result2.length()));
		in.close();
	}
	
 
//計算單個字元 乘 一個字串
	public static String multiBigIntSingle(char[] a, char b) {
		int pre = 0;
		String result = "";
		for (int j = a.length - 1; j >= 0; j--) {
			int temp = a[j] - '0';
			int tempb = b - '0';
			int sum = temp * tempb + pre;
			pre = sum / 10;
			int left = sum % 10;
			result += left;
		}
		if (pre != 0)
			result += pre;

		char[] sb = result.toCharArray();
		String value = "";
		for (int j = sb.length - 1; j >= 0; j--)
			value += sb[j];
		return value;
	}
	
 
//方法1
	public static String multiBigInt(char[] a, char[] b) {
		String c = ""; // 儲存每一行相加後的結果
		int j = 0; // 控制錯位
		for (int i = b.length - 1; i >= 0; i--) {
			c = addBigInt(c.toCharArray(), multiBigIntSingle(a, b[i]).toCharArray(), j++);
		}

		return c;
	}
	//方法二
	public static String multiBigInt(String a, String b) {

		if (a.length() == 1)
			return multiBigIntSingle(b.toCharArray(), a.charAt(0));
		if (b.length() == 1)
			return multiBigIntSingle(a.toCharArray(), b.charAt(0));

		int mid1 = a.length() / 2;
		int mid2 = b.length() / 2;

		String A = a.substring(0, mid1);
		String B = a.substring(mid1, a.length());

		String C = b.substring(0, mid2);
		String D = b.substring(mid2, b.length());

		String AC = multiBigInt(A, C);
		for (int i = 0; i < a.length() - mid1 + b.length() - mid2; i++) {
			AC += '0';
		}

		String AD = multiBigInt(A, D);
		for (int i = 0; i < a.length() - mid1; i++) {
			AD += '0';
		}

		String CB = multiBigInt(C, B);
		for (int i = 0; i < b.length() - mid2; i++) {
			CB += '0';
		}

		String BD = multiBigInt(B, D);

		String result = addBigInt(AC.toCharArray(), CB.toCharArray(), 0);
		result = addBigInt(result.toCharArray(), AD.toCharArray(), 0);
		result = addBigInt(result.toCharArray(), BD.toCharArray(), 0);

		return result;
	}
	
 
//將兩個大整數相加 len用於控制錯位相加
	public static String addBigInt(char[] a, char[] b, int len) {
		int maxlen = a.length + b.length;
		char[] revA = reverse(a);
		char[] revB = reverse(b);
		String sb = "";
		int tempa = 0;
		int tempb = 0;
		int pre = 0;
		for (int i = 0; i < maxlen; i++) {
			tempa = 0;
			tempb = 0;
			if (i < revA.length)
				tempa = revA[i] - '0';
			// 第二行要先以為len的距離 ,錯位相加
			if (i < revB.length + len && i >= len)
				tempb = revB[i - len] - '0';

			int sum = tempa + tempb + pre;
			pre = sum / 10;

			int left = sum % 10;
			sb += left;
		}
		while (pre != 0) {
			sb += pre % 10;
			pre /= 10;
		}
		char[] result = sb.toCharArray();
		String value = "";
		for (int j = result.length - 1; j >= 0; j--)
			value += result[j];
		return value;
	}
	//將一個字元陣列反轉，便於整數相加
	public static char[] reverse(char[] a) {
		char[] b = new char[a.length];

		int i = 0;
		int j = a.length - 1;
		for (; j >= 0; j--) {
			b[i] = a[j];
			i++;
		}

		return b;
	}

}

987654321 45
44444444445
44444444445

987654321 987654321
975461057789971041
975461057789971041

7415863 9874563210
73228407950200230
73228407950200230