leetcode multiply-strings

Posted 2021-02-08 zhibin123

题干：

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integerdirectly.

大意：

　　大整数乘法

输入：string

输出：string

思路：

 　　1、计算乘积中对应位的结果。不考虑进位时，根据乘数位数可以确定乘积位数，乘积对应位置的数是由乘数确定位置的数确定。如num1*num2，结果会有len1-1+len2-1+1位。计算时：

R[len - i - j]+=((num1[i]-‘0‘)*(num2[j]-‘0‘));

　

　　2、处理进位。乘积计算完后将进位由低位开始向高位进位调整每位符合十进制数

　　3、去掉前导零。

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1=num1.length(),len2=num2.length();
        int len = len1+len2-2;
        int C = 0;
        int R[220]={0};
        string ans;//对应位乘
        for(int i = 0 ; i < len1;i++){
            for(int j = 0;j < len2;j++){
                R[len - i - j]+=((num1[i]-‘0‘)*(num2[j]-‘0‘));
            }
        }
        //处理进位
        for(int i = 0 ; i < len1+len2 ;i++){
            R[i] += C;
            C = R[i]/10;
            R[i] %= 10;
        }
        //处理前导0
        int k = len1+len2;
        while(!R[k]){
            k--;
            if(k<0)
                return "0";
        }
        //压入string返回
        for(int i = k ; i >= 0;i--){
            ans.push_back(R[i]+‘0‘);
        }
        return ans;
    }
};