43、字符串相乘
题目:grey_question:：
题解:ballot_box_with_check:：
和字符串相加相同采用竖式运算思想，分析乘法的竖式运算和加法的竖式运算整体框架不变，都采用双指针，重点在于num1和num2需要选择num1循环一次和num2循环num1次。每当一个num1的元素循环相乘num2的元素后得出一个字符串结果后，左移一位（右边补个0）与上一个结果进行累加。这个累加过程和字符串相加相同。
竖式运算思想，双指针，左移补0累加得结果
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| class Solution:
def multiply(self, num1: str, num2: str) -> str:
if int(num1) == 0 or int(num2) == 0:
return "0"
i, carry = len(num1) - 1, 0
res = ""
while i >= 0:
carry = 0
str1 = ""
j = len(num2) - 1
while j >= 0:
tmp = int(num1[i]) * int(num2[j]) + carry
carry = tmp // 10
str1 = str(tmp % 10) + str1
j = j - 1
str1 = str(carry) + str1 if carry !=0 else str1
for pp in range(len(num1) - 1 - i):
str1 = str1 + "0"
res = self.addStrings(res, str1)
i = i - 1
return res
def addStrings(self, num1: str, num2: str):
res = ""
i, j, carry = len(num1) - 1, len(num2) - 1, 0
while i >= 0 or j >= 0:
n1 = int(num1[i]) if i >= 0 else 0
n2 = int(num2[j]) if j >= 0 else 0
tmp = n1 + n2 + carry
carry = tmp // 10
res = str(tmp % 10) + res
i, j = i - 1, j - 1
return "1" + res if carry == 1 else res
|
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