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43. Multiply Strings 字符串相乘

syntax padding and post digi input ola word 字符串

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:

  1. The length of both num1 and num2 is < 110.

  2. Both num1 and num2 contains only digits 0-9.

  3. Both num1 and num2 does not contain any leading zero.

  4. You must not use any built-in BigInteger library or convert the inputs to integer

    directly


123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354class Solution1: def multiply(self, num1, num2): """ :type num1: str :type num2: str :rtype: str """ if num1 is "0" or num2 is "0": return "0"
pos = [[] for i in range(len(num1) + len(num2))] startIndex = 0 for i in range(len(num2) - 1, -1, -1): curStartIndex = 0 for j in range(len(num1) - 1, -1, -1): cur = str(int(num2[i]) * int(num1[j])) curIndex = 0 for k in range
(len(cur) - 1, -1, -1): pos[startIndex + curStartIndex + curIndex].append(cur[k]) curIndex += 1 curStartIndex += 1 startIndex += 1 res = "" carry = 0 index = 0 while index < len(pos) or carry: val = carry for cur in pos[index]: val += int(cur) carry = int(val / 10) res = str(val % 10) + res index += 1 return res.lstrip("0") class Solution2: def multiply(self, num1, num2): res = 0 for i in range(len(num1)): res *= 10 n = int(num1[i]) temp_n = 0 for j in range(len(num2)): temp_n *= 10 temp_n += int(num2[j]) * n res += temp_n return str(res) s = Solution1()num1 = "999"num2 = "999"res = s.multiply(num1, num2)print(res)




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43. Multiply Strings 字符串相乘