字符串相乘
'''
今天分享的題目非常有意思：大數(shù)(字符串存儲)相乘str1*str2
m位數(shù)字*n位數(shù)字最多只有m+n位
reverse_str1=str1[::-1]
reverse_str2=str2[::-1]
tmp_res[i+j]=reverse_str1[i]*reverse_str1[j]
res[i] = tmp_res[i] % 10
tmp_res[i+1] += tmp_res[i]/10
-----------------------------------------------
input: num1, num2 = '91', '91'
tmp_res = [1,18,81,0]
res = [1,8,2,8]
'''
class Solution(object):
def multiply(self, num1, num2):
"""
:type num1: str
:type num2: str
:rtype: str
"""
lookup = {"0":0,"1":1,"2":2,"3":3,"4":4,"5":5,"6":6,"7":7,"8":8,"9":9} # 節(jié)省查找時間
if num1 == '0' or num2 == '0':
return '0'
num1, num2 = num1[::-1], num2[::-1] #將字符串逆序便于從最低位開始計算
tmp_res = [0 for i in range(len(num1)+len(num2))]
for i in range(len(num1)):
for j in range(len(num2)):
tmp_res[i+j] += lookup[num1[i]] * lookup[num2[j]]
res = [0 for i in range(len(num1)+len(num2))]
for i in range(len(num1)+len(num2)):
res[i] = tmp_res[i] % 10
if i < len(num1)+len(num2)-1: # 避免tmp_res[i+1]越界
tmp_res[i+1] += tmp_res[i]//10
return ''.join(str(i) for i in res[::-1]).lstrip('0') # 去掉最終結(jié)果頭部可能存在的‘0’
