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.
题意：字符串的乘法。
思路：模拟竖乘的思路，，跟大数计算一样，先将字符串倒置
class Solution {public:string multiply(string num1, string num2) {reverse(num1.begin(), num1.end());reverse(num2.begin(), num2.end());int len1 = num1.size(), len2 = num2.size();string ans(len1+len2+1, '0');int tmp = 0;for (int i = 0; i < len1; i++) {int cur = num1[i] – '0';tmp = 0;for (int j = 0; j < len2; j++) {int a = tmp + cur * (num2[j]-'0') + (ans[i+j]-'0');tmp = a / 10;ans[i+j] = a % 10 + '0';}int idx = len2;while (tmp != 0) {int a = tmp + (ans[i+idx]-'0');tmp = a / 10;ans[i+idx] = a % 10 + '0';idx++;}}while (!ans.empty() && ans.back() == '0')ans.pop_back();if (ans.empty()) return "0";reverse(ans.begin(), ans.end());return ans;}};
也许不是自己该去发挥的地方，还是让自己到最适合自己战斗的方面去吧!勇敢的接受自己的失败，
