Given two numbers represented as strings, return multiplication of the numbers as a string.
Note: The numbers can be arbitrarily large and are non-negative.
给定两个以字符表示的数字,求两个数字乘积,数字可以是任意大非负。以字符形式可以计算超大数相乘,不受int或long的数值范围的约束。
方法:如果给的数字不超出范围,可以直接转为整数相乘。但如果很大,就不可以直接相乘,而是第二个数的每一位和第一个数相乘然后错位相加,处理好进位和最后的数字。
Python:
class Solution(object): def multiply(self, num1, num2): """ :type num1: str :type num2: str :rtype: str """ num1, num2 = num1[::-1], num2[::-1] res = [0] * (len(num1) + len(num2)) for i in xrange(len(num1)): for j in xrange(len(num2)): res[i + j] += int(num1[i]) * int(num2[j]) res[i + j + 1] += res[i + j] / 10 res[i + j] %= 10 # Skip leading 0s. i = len(res) - 1 while i > 0 and res[i] == 0: i -= 1 return ''.join(map(str, res[i::-1]))
C++:
// Time: O(m * n) // Space: O(m + n) class Solution { public: string multiply(string num1, string num2) { const auto char_to_int = [](const char c) { return c - '0'; }; const auto int_to_char = [](const int i) { return i + '0'; }; vector<int> n1; transform(num1.rbegin(), num1.rend(), back_inserter(n1), char_to_int); vector<int> n2; transform(num2.rbegin(), num2.rend(), back_inserter(n2), char_to_int); vector<int> tmp(n1.size() + n2.size()); for(int i = 0; i < n1.size(); ++i) { for(int j = 0; j < n2.size(); ++j) { tmp[i + j] += n1[i] * n2[j]; tmp[i + j + 1] += tmp[i + j] / 10; tmp[i + j] %= 10; } } string res; transform(find_if(tmp.rbegin(), prev(tmp.rend()), [](const int i) { return i != 0; }), tmp.rend(), back_inserter(res), int_to_char); return res; } };
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