Leetcode43字符串相乘
43字符串相乘
给定两个以字符串形式表示的非负整数 num1 和 num2,返回 num1 和 num2 的乘积,它们的乘积也表示为字符串形式。
示例 1:
输入: num1 = "2", num2 = "3"
输出: "6"
示例 2:
输入: num1 = "123", num2 = "456"
输出: "56088"
说明:
- num1 和 num2 的长度小于110。
- num1 和 num2 只包含数字 0-9。
- num1 和 num2 均不以零开头,除非是数字 0 本身。
- 不能使用任何标准库的大数类型(比如 BigInteger)或直接将输入转换为整数来处理。
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/multiply-strings 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
Solution 1
按位相乘相加,未优化
执行用时:392 ms
内存消耗:13.8 MB
class Solution:
def multiply(self, num1: str, num2: str) -> str:
def strsum(num1: str, num2: str) -> str:
length = min(len(num1), len(num2))
result = ""
c = "0"
# 后面部分相加
for i in range(-1,-length-1,-1):
a = num1[i]
b = num2[i]
temp = str(int(a) + int(b) + int(c))
if len(temp) == 2:
c = "1"
else:
c = "0"
temp = temp[-1]
result = temp + result
# 前面部分与进位相加
temp = num1[:len(num1) - length] + num2[:len(num2) - length]
if c == "1":
temp = strsum(temp, c)
result = temp + result
return result
def strbitmul(num1: str, bit: str) -> str:
length = len(num1)
result = ""
c = "0"
b = bit
for i in range(-1,-length-1,-1):
a = num1[i]
temp = str(int(a) * int(b) + int(c))
if len(temp) == 2:
c = temp[0]
else:
c = "0"
temp = temp[-1]
result = temp + result
if c != "0":
result = c + result
return result
# 按位相乘相加
if num1 == "0" or num2 == "0":
return "0"
result = ""
for bit in num2:
result += "0"
temp = strbitmul(num1, bit)
result = strsum(result, temp)
return result