解题思路：
/* 4 5 6 * 1 2 3 —————————— 12 15 18 8 10 12 4 5 6 ———————————— v _ _ _ _ _ _ v 0 4 13 28 27 18 进位处理即得答案 */
1 class Solution { 2 public: 3 string multiply(string num1, string num2) { 4 int len1 = num1.length(); 5 int len2 = num2.length(); 6 vector<int>v(len1 + len2, 0); //用来存放乘积，最后的结果的位数不可能超过len1+len2 7 8 for (int i = len1 - 1; i >= 0; i--) { 9 for (int j = len2 - 1; j >= 0; j--) { 10 v[i + j + 1] += (num1[i] - '0') * (num2[j] - '0'); //此处很巧妙，对应位乘完后相加，并始终不进位 11 } 12 } 13 14 // 处理进位 15 int carry = 0; 16 for (int i = len1 + len2 - 1; i >= 0; i--) { 17 v[i] += carry; 18 carry = v[i] / 10; 19 v[i] %= 10; 20 } 21 22 string re = ""; 23 int flag = 1; 24 for (int i = 0; i < len1 + len2; ++i) { 25 if (v[i] == 0 && flag)continue;//去除前位是0的数 26 re += v[i] + '0'; 27 flag = 0; 28 } 29 if (flag) 30 re += '0';//若全为0即得0 31 return re; 32 } 33 }; 34 35 36 void T043() { 37 Solution s; 38 string num1; 39 string num2; 40 num1 = "2"; 41 num2 = "3"; 42 cout << s.multiply(num1, num2) << endl; 43 num1 = "123"; 44 num2 = "456"; 45 cout << s.multiply(num1, num2) << endl; 46 num1 = "0"; 47 num2 = "456"; 48 cout << s.multiply(num1, num2) << endl; 49 }