| 标题： | Multiply Strings |
| 通过率： | 20.8% |
| 难度： | 中等 |
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.
本题是模拟大数字的相乘，如果用java的bigInteger去处理的话就没有本题的实际意义了，本题想要做的就是模拟乘法，乘法如何计算很容易判断，
难度就是模拟计算，用一个数组去储存临时的数据，然后一个进位缓存，去保存进位，本题难度就是位置的确定。
直接看代码如何处理：
1 public class Solution { 2 public String multiply(String num1, String num2) { 3 int len1=num1.length(),len2=num2.length(),numTemp1=0,numTemp2=0,carry=0,sum=0; 4 String result=""; 5 if(len1==0||len2==0)return "0"; 6 int [] tmpArray=new int[len1+len2]; 7 int location=0; 8 for(int i=0;i<len1;i++){ 9 numTemp1=Integer.valueOf(num1.charAt(len1-i-1)-'0'); 10 carry=0; 11 12 for(int j=0;j<len2;j++){ 13 numTemp2=Integer.valueOf(num2.charAt(len2-j-1)-'0'); 14 sum=numTemp1*numTemp2+tmpArray[len1+len2-j-i-1]+carry; 15 carry=sum/10; 16 sum=sum%10; 17 tmpArray[len1+len2-j-i-1]=sum; 18 location=j; 19 } 20 if(carry!=0)tmpArray[len1+len2-i-location-2]+=carry; 21 22 } 23 int start=0; 24 for(;start<len1+len2;start++){ 25 if(tmpArray[start]!=0)break; 26 } 27 if(start==(len1+len2))return "0"; 28 for(;start<len1+len2;start++){ 29 result+=tmpArray[start]; 30 31 } 32 return result; 33 34 35 } 36 }
python代码：
1 class Solution: 2 def multiply(self, num1, num2): 3 num1, num2 = num1[::-1], num2[::-1] 4 result = [0 for i in range(len(num1) + len(num2))] 5 for i in range(len(num1)): 6 for j in range(len(num2)): 7 result[i + j] += int(num1[i]) * int(num2[j]) 8 carry, total = 0, [] 9 for digit in result: 10 sum = carry + digit 11 carry = sum / 10 12 total.insert(0, str(sum % 10)) 13 while len(total) > 1 and total[0] == "0": 14 del total[0] 15 return ''.join(total)