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Problem Definition:
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.
Solution:
大数乘法，常用的算法有：Long Multiplication、分治、FFT.
1）逐位乘。
1 # @param {string} num1 2 # @param {string} num2 3 # @return {string} 4 def multiply(self, num1, num2): 5 if num1==‘0‘ or num2==‘0‘: 6 return ‘0‘ 7 num1,num2=map(int, num1), map(int, num2) 8 n1,n2=len(num1), len(num2) 9 result=[0]*(n1+n2) 10 for i in range(n1-1, -1, -1): 11 carry=0 12 for j in range(n2-1, -1, -1): 13 pz=(n1-1-i)+(n2-1-j) 14 result[pz]+=carry+num1[i]*num2[j] 15 carry=result[pz]/10 16 result[pz]%=10 17 result[n1-1-i+n2]+=carry 18 19 while result[-1]==‘0‘: 20 result.pop() 21 return ‘‘.join(map(str, result[::-1]))
2）分组乘。64位的二进制数可以表达19位的十进制数，因此两个9位十进制数的乘积可以用64位二进制数存放而不溢出。
因此可以把用数组表示的大整数，每9位为一组，然后group-by-group地相乘，比digit-by-digit会快些。要预处理，对数位分组。
1 # @param {string} num1 2 # @param {string} num2 3 # @return {string} 4 def multiply(self, num1, num2): 5 # convert number string into list of 9-digit numbers 6 def _convert2List(numString): 7 numList = [0] * ((len(numString) + 8) // 9) 8 # store 9-digit number in reverse order 9 numList[0] = int(numString[-9:]) 10 for index in range(1, len(numList)): 11 numList[index] = int(numString[-(index+1)*9 : -index*9]) 12 return numList 13 14 # multiply a number list with a at most 9-digit multiplier 15 def _multiplySingle(result, start, numList, multiplier): 16 carry = 0 17 for index in range(len(numList)): 18 resultIndex = index + start 19 carry += (numList[index] * multiplier) + result[resultIndex] 20 carry, result[resultIndex] = divmod(carry, 1000000000) 21 result[len(numList) + start] = carry 22 23 if num1 == ‘0‘ or num2 == ‘0‘: 24 return ‘0‘ 25 # convert string to integer 26 num1List, num2List = _convert2List(num1), _convert2List(num2) 27 result = [0] * ((len(num1) + len(num2) + 16) // 9) 28 for index in range(len(num2List)): 29 _multiplySingle(result, index, num1List, num2List[index]) 30 31 # remove leading 0s 32 while result[-1] == 0: 33 result.pop() 34 # convert to string 35 return str(result[-1]) + ‘‘.join(map(lambda x : string.zfill(x, 9), result[:-1][::-1]))
实际上直接写
return str(int(num1)*int(num2))
也能通过，因为Python本身就有处理溢出的机制。int会被自动转换成long，而long是大整数类型，限制其长度的是内存大小。
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原文地址：http://www.cnblogs.com/acetseng/p/4702500.html
