Multiply Strings 題解

題目來源：https://leetcode.com/problems/multiply-strings/description/

Description

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

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1
   and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

Solution

class Solution {
private:
    string multiOneNum(const string& nums, char num) {
        int overflow = 0;
        string res;
        int size = nums.size();
        for
 
 (int i = size - 1; i >= 0; i--) {
            int temp = (nums[i] - '0') * (num - '0');
            res.push_back(static_cast<char>((temp + overflow) % 10 + '0'));
            overflow = (temp + overflow) / 10;
        }
        if (overflow > 0)
            res.push_back(static_cast
 
<char>(overflow + '0'));
        return string(res.rbegin(), res.rend());
    }
    string addTwoNums(const string& num1, const string& num2) {
        int overflow = 0;
        string res;
        int i, j;
        for (i = num1.length() - 1, j = num2.length() - 1;
                i >= 0 && j >= 0; --i, --j) {
            int temp = (num1[i] - '0') + (num2[j] - '0');
            res.push_back(static_cast<char>((temp + overflow) % 10 + '0'));
            overflow = (temp + overflow) / 10;
        }
        while (i >= 0) {
            int temp = num1[i--] - '0';
            res.push_back(static_cast<char>((temp + overflow) % 10 + '0'));
            overflow = (temp + overflow) / 10;
        }
        while (j >= 0) {
            int temp = num2[j--] - '0';
            res.push_back(static_cast<char>((temp + overflow) % 10 + '0'));
            overflow = (temp + overflow) / 10;
        }
        if (overflow > 0)
            res.push_back(static_cast<char>(overflow + '0'));
        while (res.back() == '0')
            res.pop_back();
        return string(res.rbegin(), res.rend());
    }
public:
    string multiply(string num1, string num2) {
        string& longNum = num1.length() >= num2.length() ?
                          num1 : num2;
        string& shortNum = num1.length() < num2.length() ?
                           num1 : num2;
        int shortLen = shortNum.length();
        string res = "0";
        if (shortLen == 1 && shortNum[shortLen - 1] == '0')
            return res;
        for (int i = shortLen - 1; i >= 0; i--) {
            if (shortNum[i] == '0')
                continue;
            string temp = multiOneNum(longNum, shortNum[i]);
            int zeroCount = shortLen - 1 - i;
            while (zeroCount--) {
                temp.push_back('0');
            }

            if (res == "0")
                res = temp;
            else
                res = addTwoNums(res, temp);
        }
        return res;
    }
};

解題描述

這道題題意是對求出兩個用字符串表示的數字相乘結果的字符串表示。上面是我個人的解法，使用的是類似手算乘法的思想，但是在結果字符串上需要多次反轉，會大大增加時間復雜度。

更優解法

評論區給出了一種也是類似於手算的解法，但不需要對結果字符串進行反轉，原因在於該解法先預留了充足的結果串長度：

class Solution {
public:
    string multiply(string num1, string num2) {
        int size1 = num1.size(), size2 = num2.size();
        string res(size1 + size2, '0');
        int i, j, carry; // carry即為進位
        for (i = size1 - 1; i >= 0; --i) {
            carry = 0;
            for (j = size2 - 1; j >= 0; --j) {
                int temp = (res[i + j + 1] - '0') +
                        (num1[i] - '0') * (num2[j] - '0') + carry;
                res[i + j + 1] = temp % 10 + '0';
                carry = temp / 10;
            }
            res[i] += carry; // 加上最後的進位
        }
        int start = res.find_first_not_of('0');
        if (start == string::npos)
            return "0";
        else
            return res.substr(start);
    }
};

[Leetcode] Multiply Strings