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.
这道题让我们求两个字符串数字的相乘,输入的两个数和返回的数都是以字符串格式储存的,这样做的原因可能是这样可以计算超大数相乘,可以不受int或long的数值范围的约束,那么我们该如何来计算乘法呢,我们小时候都学过多位数的乘法过程,都是每位相乘然后错位相加,那么这里就是用到这种方法,参见网友JustDoIt的博客,把错位相加后的结果保存到一个一维数组中,然后分别每位上算进位,最后每个数字都变成一位,然后要做的是去除掉首位0,最后把每位上的数字按顺序保存到结果中即可,代码如下:
class Solution { public: string multiply(string num1, string num2) { string res; int n1 = num1.size(), n2 = num2.size(); int k = n1 + n2 - 2, carry = 0; vector<int> v(n1 + n2, 0); for (int i = 0; i < n1; ++i) { for (int j = 0; j < n2; ++j) { v[k - i - j] += (num1[i] - '0') * (num2[j] - '0'); } } for (int i = 0; i < n1 + n2; ++i) { v[i] += carry; carry = v[i] / 10; v[i] %= 10; } int i = n1 + n2 - 1; while (v[i] == 0) --i; if (i < 0) return "0"; while (i >= 0) res.push_back(v[i--] + '0'); return res; } };