LeetCode——043

/*
43. Multiply Strings My Submissions QuestionEditorial Solution
Total Accepted: 59659 Total Submissions: 255286 Difficulty: Medium
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.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.
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*/
/*
解题思路：
具体步骤在代码中已经写的比较清楚了。本题的思想就是一个数的第i位数与另一个数的第j位相乘，其相乘的结果一定放在结果数中的i+j-1处。最后做进位处理就可以了。

*/

class Solution {
public:
    string multiply(string num1, string num2) {

        //大数相乘（之前写过，就直接搬代码了）

        string res="";
    //两个数的位数
    int m = num1.size(), n = num2.size();

    //一个i位数乘以一个j位数，结果至少是i+j-1位数
    vector<long long> tmp(m + n - 1);

    //每一位进行笛卡尔乘法
    for (int i = 0; i < m; i++){
         int a = num1[i] - '0';
        for (int j = 0; j < n; j++){
            int b = num2[j] - '0';
            tmp[i + j] += a*b;
        }
    }
    //进行进位处理，注意左侧是大右侧是小
    int carry = 0;
    for (int i = tmp.size() - 1; i >= 0; i--){
        int t = tmp[i] + carry;
        tmp[i] = t % 10;
        carry = t /10;
    }
    //若遍历完仍然有进位
    while (carry != 0){
        int t = carry % 10;
        carry /= 10;
        tmp.insert(tmp.begin(), t);
    }
    //将结果存入到返回值中
    for (auto a : tmp){
        res = res + to_string(a);
    }
    if(res.size()>0&&res[0]=='0')return "0";
    return res;

    }
};