力扣算法题—043字符串相乘

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
*/

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1 = num1.length();
        int len2 = num2.length();
        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
        
        for (int i = len1 - 1; i >= 0; i--) {
            for (int j = len2 - 1; j >= 0; j--) {
                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
            }
        }

        // 处理进位
        int carry = 0;
        for (int i = len1 + len2 - 1; i >= 0; i--) {
            v[i] += carry;
            carry = v[i] / 10;
            v[i] %= 10;
        }

        string re = "";
        int flag = 1;
        for (int i = 0; i < len1 + len2; ++i) {
            if (v[i] == 0 && flag)continue;//去除前位是0的数
            re += v[i] + '0';
            flag = 0;
        }
        if (flag)
            re += '0';//若全为0即得0
        return re;
    }
};


void T043() {
    Solution s;
    string num1;
    string num2;
    num1 = "2";
    num2 = "3";
    cout << s.multiply(num1, num2) << endl;
    num1 = "123";
    num2 = "456";
    cout << s.multiply(num1, num2) << endl;
    num1 = "0";
    num2 = "456";
    cout << s.multiply(num1, num2) << endl;
}