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.
思路:
大整数运算的思路,模拟乘法运算
代码:
class Solution { public: string multiply(string num1, string num2) { int len1 = num1.size(), len2 = num2.size(), len = len1 + len2; string str(len, '0'); for (int i = len1 - 1; i >= 0; i--) { int a = num1[i] - '0'; for (int j = len2 - 1, k = len2 + i; j >= 0; j--, k--) { int b = num2[j] - '0'; int c = str[k] - '0'; int t = b * a + c; str[k] = t % 10 + '0'; int d = (str[k-1] - '0') + t / 10; if (d >= 10) //开始这里没有等号,检查了很久才发现,细心啊细心 str[k-2] = str[k-2] - '0' + d / 10 + '0'; str[k-1] = d % 10 + '0'; } } int x = 0; while (str[x] == '0') x++; if (str.substr(x, len - x) == "") return "0"; return str.substr(x, len - x); } };