数据结构与算法实践之路

大数乘法

2017-05-10  本文已影响0人  HangChen

大数乘法:

public class BigNumberMulti {  
    public static void main(String[] args) {  
        String num1Str = "24566";  
        String num2Str = "452053";  
        int[] num1 = new int[num1Str.length()];  
        int[] num2 = new int[num2Str.length()];  
        //将字符串转换为整数数组,注意,24566应转化成{6,6,5,4,2}  
        for(int i = 0; i < num1Str.length();i++) {  
            num1[num1Str.length() -1- i] = num1Str.charAt(i)-'0';  
        }  
        for(int i = 0; i < num2Str.length();i++) {  
            num2[num2Str.length() -1- i] = num2Str.charAt(i)-'0';  
        }  
        int[] result = multiply(num1, num2);  
        for(int i = result.length-1; i>=0;i--){  
            System.out.print(result[i]);  
        }  
    }  
  
    public static int[] multiply(int[] num1, int[] num2) {  
        int lengthOfNum1 = num1.length;  
        int lengthOfNum2 = num2.length;  
        // 分配一个空间,用来存储运算的结果,n1长的数* n2长的数,结果不会超过n1+n2长  
        int[] result = new int[lengthOfNum1 + lengthOfNum2];  
        // 先不考虑进位问题,根据小学竖式的乘法运算,n1的第i位与n2的第j位相乘,结果应该存放在结果的第i+j位上  
        for (int i = 0; i < lengthOfNum1; i++) {  
            for (int j = 0; j < lengthOfNum2; j++) {  
                result[i + j] += num1[i] * num2[j];  
            }  
        }  
        // 单独处理进位问题  
        for (int i = 0; i < lengthOfNum1 + lengthOfNum2 -1; i++) {  
            if (result[i] > 10) {  
                result[i + 1] += result[i] / 10;  
                result[i] %= 10;  
            }  
        }  
          
        return result;  
  
    }  
}  
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