LeetCode#43 Multiply Strings

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。

    # @param {string} num1
    # @param {string} num2
    # @return {string}
    def multiply(self, num1, num2):
        if num1=='0' or num2=='0':
            return '0'
        num1,num2=map(int, num1), map(int, num2)
        n1,n2=len(num1), len(num2)
        result=[0]*(n1+n2)
        for i in range(n1-1, -1, -1):
            carry=0
            for j in range(n2-1, -1, -1):
                pz=(n1-1-i)+(n2-1-j)
                result[pz]+=carry+num1[i]*num2[j]
                carry=result[pz]/10
                result[pz]%=10
            result[n1-1-i+n2]+=carry
        19         while result[-1]=='0':
            result.pop()
        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。

    # @param {string} num1
    # @param {string} num2
    # @return {string}
    def multiply(self, num1, num2):
        # convert number string into list of 9-digit numbers
        def _convert2List(numString):
            numList = [0] * ((len(numString) + 8) // 9)
            # store 9-digit number in reverse order
            numList[0] = int(numString[-9:])
            for index in range(1, len(numList)):
                numList[index] = int(numString[-(index+1)*9 : -index*9])
            return numList

        # multiply a number list with a at most 9-digit multiplier
        def _multiplySingle(result, start, numList, multiplier):
            carry = 0
            for index in range(len(numList)):
                resultIndex = index + start
                carry += (numList[index] * multiplier) + result[resultIndex]
                carry, result[resultIndex] = divmod(carry, 1000000000)
            result[len(numList) + start] = carry

        if num1 == '0' or num2 == '0':
            return '0'
        # convert string to integer
        num1List, num2List = _convert2List(num1), _convert2List(num2)
        result = [0] * ((len(num1) + len(num2) + 16) // 9)
        for index in range(len(num2List)):
            _multiplySingle(result, index, num1List, num2List[index])

        # remove leading 0s
        while result[-1] == 0:
            result.pop()
        # convert to string
        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。