[LeetCode]题解（python）：043-Multiply Strings

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题目来源

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https://leetcode.com/problems/multiply-strings/

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.

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题意分析

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Input: two numbers expressed as string

Output:the multiply of the two sums

Conditions:数可以无限大，做两个数的乘法

如："23650379502752" 和 "865382861454"

结果："20466633088564555427721408"

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题目思路

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首先将两个str转化为整数的list，然后考虑到乘积的位数必然小于等于len（str1）+len（str2），先初始化乘积为0的list，然后按照位数相乘的规律去做

注意：

1 最后结果需要将大数的0去掉，同时如果结果为0需要返回串“0”

2 翻转：mul.reverse()

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AC代码（Python）

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_author_ = "YE"
# -*- coding:utf-8 -*-

class Solution(object):
    def multiply(self, num1, num2):
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        len1 = len(num1)
        len2 = len(num2)

        list1 = [0 for i in range(len1)]
        list2 = [0 for i in range(len2)]

        for i in range(len1):
            list1[len1 - 1 - i] = int(num1[i])
        for i in range(len2):
            list2[len2 - 1 -i] = int(num2[i])

        # print(list1,list2)

        mul = [0 for i in range(len1 + len2)]

        for i in range(len2):
            carry = 0
            for j in range(len1):
                mul[i + j] = mul[i + j] + carry + (list2[i] * list1[j]) % 10

                carry = (list2[i] * list1[j]) // 10

                if mul[i + j] >= 10:
                    carry = carry + mul[i + j] // 10
                    mul[i + j] = mul[i + j] % 10


            if carry > 0:
                mul[i + len1] += carry
                if mul[i + len1] > 10:
                    mul[i + len1] = mul[i + len1] % 10
                    carry += mul[i + len1] // 10

        index = len1 + len2 - 1
        while index >= 0:
            if mul[index] > 0:
                break
            index -= 1

        if index + 1 < len1 + len2:
            mul[index+1:] = []

        mul.reverse()

        s = ''
        for i in range(len(mul)):
            s += str(mul[i])
        if s == '':
            s = '0'
        return s

str1 = '12312'
str2 = '4234'
s = Solution()

print(s.multiply(str1,str2))