Python世界:力扣题43大数相乘算法实践
任务背景
问题来自力扣题目43:字符串相乘,大意如下:
Given two non-negative integers
num1
andnum2
represented as strings, return the product ofnum1
andnum2
, also represented as a string.
翻译下,需求是:实现大数相乘,字符串乘法
- 输入为非负整数两个字符串
- 要求输出该大数值的乘积
思路分析
方案1
自然的想法是,模拟乘法运算,考验对实际问题的计算机转换,先手动模拟下计算过程,提炼其中算法,如果最高位相乘及低位相加无累进,则提前退出。
shell
99*99=9801 2*2=4
10*10=100 2*2=3
以下示例,运行时间击败32%:
python
# sol1:暴力法遍历
class Solution:
def multiply(self, num1: str, num2: str) -> str:
# corner case
if num1 == "0" or num2 == "0":
return "0"
elif num1 == "1":
return num2
elif num2 == "1":
return num1
# common case
len1 = len(num1)
len2 = len(num2)
len_sum = len1 + len2
len_max = max(len1, len2)
len_min = min(len1, len2)
# 从低位往高位相互进位,个、十、百、千、......
n1_rev = num1[::-1]
n2_rev = num2[::-1]
multi_res_list = []
base = 10
c = 0 # carrier
# 暴力法
for b in range(len_sum + 1):
val = 0
res = 0
# 获取一个阶的结果,如百、十、千
for i in range(len1):
if i > b or b - i >= len2: # i,j比目标进位大,已到头
continue
j = b - i # j>=0 && j<len2
n1 = int(n1_rev[i])
n2 = int(n2_rev[j])
res += n1 * n2
# 处理一个阶的结果
res += c
c = res // base
val = res - c * base
assert(val < base)
if (c == 0 and val == 0 and b > len_max): # 去除冗余前导零
continue
multi_res_list.append(str(val))
# 将列表逆序并转化为字符串输出
# multi_res_list = multi_res_list.reverse() # 未按预期运行,输出结果为None
multi_res_list = multi_res_list[::-1]
non_zero_idx = 0
for val in multi_res_list:
if (val == "0"):
non_zero_idx += 1
else:
break
multi_res_list = multi_res_list[non_zero_idx:]
multi_res_str = "".join(multi_res_list) # 列表转字符串
return multi_res_str
方案2
尝试进一步改进:
- 通过限制上下界,降低内外for循环次数
- 内循环len1选两者较小的长度
- 如果i大于b时,直接break
- 外循环b设计提前退出条件,当前导都是零时,无计算必要
- 不整体逆序,直接从末尾字符低位往高位移动(TBD)
python
# sol2:beat 42.5%
class Solution:
def multiply(self, num1: str, num2: str) -> str:
# corner case
if num1 == "0" or num2 == "0":
return "0"
elif num1 == "1":
return num2
elif num2 == "1":
return num1
# common case
len1 = len(num1)
len2 = len(num2)
len_sum = len1 + len2
len_max = max(len1, len2)
len_min = min(len1, len2)
# 从低位往高位相互进位,个、十、百、千、......
n1_rev = num1[::-1]
n2_rev = num2[::-1]
multi_res_list = []
base = 10
c = 0 # carrier
# 暴力法
for b in range(len_sum): # b [0, len_sum-1]
val = 0
res = 0
# 获取一个阶的结果,如百、十、千
for i in range(len1):
if i > b:
break
if b - i >= len2: # i,j比目标进位大,已到头
continue
j = b - i # j>=0 && j<len2
n1 = int(n1_rev[i])
n2 = int(n2_rev[j])
res += n1 * n2
# 处理一个阶的结果
res += c
c = res // base
val = res - c * base
assert(val < base)
if (c == 0 and val == 0 and b > len_max): # 去除冗余前导零
continue
multi_res_list.append(str(val))
if (b + 1 == len_sum and c == 0):
break # 最高位相乘无进位
# 将列表逆序并转化为字符串输出
# multi_res_list = multi_res_list.reverse() # 未按预期运行,输出结果为None
multi_res_list = multi_res_list[::-1]
non_zero_idx = 0
for val in multi_res_list:
if (val == "0"):
non_zero_idx += 1
else:
break
multi_res_list = multi_res_list[non_zero_idx:]
multi_res_str = "".join(multi_res_list) # 列表转字符串
return multi_res_str
方案3
网上参考的一种实现,运行时间对比:
python
# # sol3:beat 29.9%
# # 参考解法:https://blog.csdn.net/huqinweI987/article/details/88797663
class Solution:
def multiply(self, num1: str, num2: str) -> str:
if num1 == '0' or num2 == '0':#有0就不用乘了。
return '0'
res = ''
carry = 0#初始化
# 两个数的长度,分别都减1
m = len(num1) - 1
n = len(num2) - 1
# m和n都是len减1,是因为,15*15中,不算被动进位,能用来主动计算乘法的,最高位就是百位,10*10=100,是主动计算的最高位。
# k就在[0,m+n]的区间:代表主动计算乘法的位(最后多出来的进位单独给出)。k=0,i和j都是0,5*5,对应个位结果。
# k=1,i和j分别是0、1和1、0组合,是10和5或者5和10,对应十位的结果,
# k=2,i和j分别是1、1(其他组合不满足筛选条件,我计算的就是百位,不能把5也拿来用吧,把乘法写一下就出来了),代表10和10相乘,对应百位结果。
for k in range(m + n + 1):
print('k:',k)
# i是所有输出位,包括k=m+n,不包括m+n+1,其实就是遍历所有可能的num1和num2的单独一位,做一个总的累加
# i、j他俩是严格针对k的互补关系。i = 时,j = 1;i = 1时,j = 0,他们都对应结果的"下标"k=1,也就是"十位"
sum = carry#先把进位计算进来(这个顺序其实无所谓,但是如果不是先进位,就要给sum清零了)
for i in range(k + 1):#k其实就是结果位。i和j是根据k做的互补,严格对应一个结果底位。
j = k - i
if i <= m and j <= n:
index_i = m - i# 转换,字符串形式,i=0其实代表的是最大的那个数,不是最小的,index_i才是最小的数。
index_j = n - j
sum += int(num1[index_i]) * int(num2[index_j])#
# 拼接结果字符串,遍历完当前k对应的所有i和j的组合,当前位的结果已经出炉,可以拼接了。比如15*15的最后一位5*5,是由当前位停留结果5和进位2组成的,当前结果就留在这。
res = str(sum % 10) + res#从低位向高位迭代,使用新的sum模,后加res的拼接方式。
carry = sum // 10#进位,5*5=25,进位2
if carry:#最后一位了,k迭代的是乘法计算,当然可能发生进位,比如33*44中,k是0到2,最高位3*4肯定要进位的
res = str(carry) + res
return res
方案4
参考烧脑版的第二个直观优雅的思路,进行python实现:先乘,再进位,代码如下:
python
# sol4:beat 34%
# 参考烧脑版的第二个思路进行python实现:先乘,再进位
class Solution:
def multiply(self, num1: str, num2: str) -> str:
# corner case
if num1 == "0" or num2 == "0":
return "0"
elif num1 == "1":
return num2
elif num2 == "1":
return num1
# common case
len1 = len(num1)
len2 = len(num2)
len_sum = len1 + len2
# 从低位往高位相互进位,个、十、百、千、......
n1_rev = num1[::-1]
n2_rev = num2[::-1]
multi_res_list = []
# 整体乘完放1个数组
num_rev_list = [0]*(len_sum)
for i in range(len1):
for j in range(len2):
n1 = int(n1_rev[i])
n2 = int(n2_rev[j])
num_rev_list[i + j] += n1 * n2
base = 10
# 统一处理进制问题
for i in range(len(num_rev_list) - 1):
num_rev_list[i+1] += num_rev_list[i] // base
num_rev_list[i] = num_rev_list[i] % base
# 处理最高位的corner case
# if (num_rev_list[len_sum - 1] == 0):
multi_res_list = num_rev_list[::-1]
non_zero_idx = 0
for val in multi_res_list:
if (val == 0):
non_zero_idx += 1
else:
break
multi_res_list = multi_res_list[non_zero_idx:]
multi_res_str = "".join((str(i) for i in multi_res_list)) # 列表中的每个数字转字符串
return multi_res_str
无测试套主调
无测试套版本主调:
python
# 无测试套版本主调
if __name__ == '__main__':
print('start!')
# num1 = "2"
# num2 = "3"
# ret = "6"
# num1 = "99"
# num2 = "99"
# ret = "9801"
num1 = "10"
num2 = "10"
ret = "100"
# num1 = "1"
# num2 = "123456789"
# ret = "123456789"
# num1 = "123456789"
# num2 = "0"
# ret = "0"
# num1 = "123"
# num2 = "456"
# ret = "56088"
# num1 = "37689269854"
# num2 = "12548698156"
# ret = "472951271117876189224"
# num1 = "6"
# num2 = "501"
# ret = "3006"
problem = Solution()
res = problem.multiply(num1, num2)
assert(res == ret)
print(res, "right!")
print('done!')
测试套主调
编写含单元测试的主调:
python
# 导入单元测试
import unittest
# function...
# 编写测试套
class TestSol(unittest.TestCase):
def test_bound1(self):
num1 = "2"
num2 = "3"
ret = "6"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_bound2(self):
num1 = "37689269854"
num2 = "12548698156"
ret = "472951271117876189224"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_bound3(self):
num1 = "1"
num2 = "123456789"
ret = "123456789"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_bound4(self):
num1 = "123456789"
num2 = "0"
ret = "0"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_special1(self):
num1 = "6"
num2 = "501"
ret = "3006"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_special2(self):
num1 = "10"
num2 = "10"
ret = "100"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_special3(self):
num1 = "99"
num2 = "99"
ret = "9801"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
def test_common_case(self):
num1 = "123"
num2 = "456"
ret = "56088"
sol = Solution()
self.assertEqual(sol.multiply(num1, num2), ret)
# 含测试套版本主调
if __name__ == '__main__':
print('start!')
unittest.main() # 启动单元测试
print('done!')
本文小结
为便于深入理解进制转换和乘法原理,同时提高编程能力,demo程序中新增单元测试代码实现。
卡壳点:
- 陷入复杂算法细节,而不是以终为始。在没明确思路前,先实现再优化,用笨办法/暴力法解决了,再尝试改进。
- corner case处理不当。 结尾中,输出字符串前导0场景。 中间乘积结果为0,进位符为0场景未考虑周全。
总的来说,推荐solution4方法进行解题。
此外,进阶想一想,如果将其变成大数加法,这个程序能否只改两三行代码,即可输出正确结果?再如,改成八进制乘法,如何搞?
题解参考
- 暴力法分解为单数相乘:LeetCode:43 multiply 大数乘法的数学直观理解
- 暴力法分解为字符串相加:字符串相乘(大数相乘、相加)
- 进阶烧脑版高效算法实现:博客园版本、CSDN版本
涉及知识点