Multiply Strings

Multiply Strings

问题：

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

思路：

　　手工模拟乘法

我的代码：

public class Solution {

    public String multiply(String num1, String num2) {

        if(num1==null || num1.length()==0 || num2==null || num2.length()==0 || num1.charAt(0)=='0' || num2.charAt(0)=='0')    return "0";

        if(num1.charAt(0) == '0') num1 = num1.substring(1);

        if(num2.charAt(0) == '0') num2 = num2.substring(1);

        int one = num1.length();

        int two = num2.length();

        String[] rst = new String[one];

        

        for(int i=one-1; i>=0; i--)

        {

            StringBuffer sb = new StringBuffer();

            for(int j=one-1; j>i; j--) sb.append('0');

            int a = (int)(num1.charAt(i)-'0');

            int carry = 0;

            for(int j=two-1; j>=0; j--)

            {

                int b = (int)(num2.charAt(j)-'0');

                int c = a*b + carry;

                sb.append(c%10);

                carry = c/10;

            }

            if(carry != 0)  sb.append(carry);

            rst[i] = sb.toString();

        }

        

        int maxLength = Integer.MIN_VALUE;

        for(String s: rst)  maxLength = Math.max(maxLength, s.length());

        StringBuffer sb = new StringBuffer();

        int carry = 0;

        for(int i=0; i<maxLength; i++)

        {

            int sum = carry;

            for(String s: rst)

            {

                sum += (s.length()<=i?0:(int)(s.charAt(i)-'0'));

            }

            sb.append(sum%10);

            carry = sum/10;

        }

        if(carry != 0) sb.append(carry);

        return sb.reverse().toString();

        

    }

    

}

View Code

他人代码：

public class Solution {

    public String multiply(String num1, String num2) {

        if (num1 == null || num2 == null) {

            return null;

        }

        

        int len1 = num1.length(), len2 = num2.length();

        int len3 = len1 + len2;

        int i, j, product, carry;



        int[] num3 = new int[len3];

        for (i = len1 - 1; i >= 0; i--) {

            carry = 0;

            for (j = len2 - 1; j >= 0; j--) {

                product = carry + num3[i + j + 1] +

                    Character.getNumericValue(num1.charAt(i)) *

                    Character.getNumericValue(num2.charAt(j));

                num3[i + j + 1] = product % 10;

                carry = product / 10;

            }

            num3[i + j + 1] = carry;

        }



        StringBuilder sb = new StringBuilder();

        i = 0;



        while (i < len3 - 1 && num3[i] == 0) {

            i++;

        }



        while (i < len3) {

            sb.append(num3[i++]);

        }



        return sb.toString();

    }

}

View Code

学习之处：

• 记住一个结论：两个数相乘，位数不会大于二者的位数之和，所以这样可以提前设定结果数组的大小了，不用再像我用了string的数组去存储中间结果，浪费空间，浪费时间
• 尽量少的用a,b等命名，用product多大气啊，另外能直接运算的就直接运算把，别再用变量进行记录传递了，关键是没作用，看起来代码啰嗦。