[force buckle]43 String multiplication

The original way

class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        if num1 == '0' or num2 == '0':return '0'
        sgn = (num1[0] == '-') ^ (num2[0] == '-')
        num1 = list(num1[::-1])
        num2 = list(num2[::-1])
        if num1[-1] == '-':num1.pop()
        if num2[-1] == '-':num2.pop()
        num1 = list(map(int,num1))
        num2 = list(map(int,num2))
        res = [0] * 1000
        t = 0
        for i in range(len(num1)):
            for j in range(len(num2)):
                res[i + j] += num1[i] * num2[j] + t
                t = res[i + j] // 10
                res[i + j] %= 10
            res[i + len(num2)] += t
            t = 0
        res = list(map(str,res))
        while res[-1] == '0': res.pop()
        if sgn:res.append('-')
        return ''.join(res[::-1])

Improved practice

class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        if num1 == '0' or num2 == '0':return '0'
        sgn = (num1[0] == '-') ^ (num2[0] == '-')
        num1 = list(num1[::-1])
        num2 = list(num2[::-1])
        if num1[-1] == '-':num1.pop()
        if num2[-1] == '-':num2.pop()
        num1 = list(map(int,num1))
        num2 = list(map(int,num2))
        res = [0] * 1000
        for i in range(len(num1)):
            for j in range(len(num2)):
                res[i + j] += num1[i] * num2[j]
        for i in range(len(num1) + len(num2)):
            res[i + 1] += res[i] // 10
            res[i] %= 10
        res = list(map(str,res))
        while res[-1] == '0': res.pop()
        if sgn:res.append('-')
        return ''.join(res[::-1])

The original method is a simple simulation of integer multiplication in primary school , And the code of the improved version , Separate the process of bit by bit multiplication from the process of carry , Although there is little difference in code length , But the difficulty of writing the code is slightly reduced .
As shown in the figure .

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