LeetCode 43. Multiply Strings

[woodyiiiiiii]

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 integer directly.

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大数相乘。乘法本质上就是移位的加法，所以我的思路是第一个字符串对第二个字符串的每位字符相乘，左移，将得到的结果相加就是答案。
我的做法如下：

class Solution {
public:
    string multiply(string num1, string num2) {
        string res;
        int n2 = num2.length();
        for (int i = n2 - 1; i >= 0; --i) {
            string s = singleMul(num1, num2[i] - '0', n2 - i - 1);
            res = strAdd(res, s);
        }
        
        // 判断字符串是否为全0
        for (int i = 0; i < res.length(); ++i) {
            if (res[i] != '0') return res;
        }
        return "0";
    }
    //相乘函数
    string singleMul(string num, int k, int p) {
        int n1 = num.length();
        int up = 0;
        for (int i = n1 - 1; i >= 0; --i) {
            char ch = num[i];
            int aInt = ch - '0';
            ch = (aInt * k + up) % 10 + '0';
            up = (aInt * k + up) / 10;
            num[i] = ch;
        }
        //如果还有进位，则要插入到字符串最前
        if (up) {
            num.insert(0, 1, up + '0');
        }
        while (p > 0) {
            num.push_back('0');
            --p;
        }
        return num;
    }
    //字符串相加函数
    string strAdd(string res, string s) {
        string result;
        int resLen = res.length(), sLen = s.length();
        if (resLen == 0) return s;
        int i = resLen - 1, j = sLen - 1;
        int up = 0;
        while (i >= 0 && j >= 0) {
            int a = (res[i] - '0') + (s[j] - '0') + up;
            result.push_back(a % 10 + '0');
            up = a / 10;
            --i;
            --j;
        }
        while (i >= 0) {
            char ch = (res[i] - '0' + up) % 10 + '0';
            up = (res[i] - '0' + up) / 10;
            result.push_back(ch);
            --i;
        }
        while (j >= 0) {
            char ch = (s[j] - '0' + up) % 10 + '0';
            up = (s[j] - '0' + up) / 10;
            result.push_back(ch);
            --j;
        }

        //不要忘了也要算进位
        if (up != 0) result.push_back(up + '0');
        i = 0;
        j = result.length() - 1;
        //因为顺序是反着的，所以字符串要翻转一遍
        while (i < j) {
            char temp = result[i];
            result[i] = result[j];
            result[j] = temp;
            ++i;
            --j;
        }
        return result;
    }
};

遇到的困难点：

• Java对字符串更改某个字符很麻烦，不如C++的STL方便
• C++的string类的函数，比如insert, pushback等
• 相加函数中要注意进位和最后的反转(其实可以前插就不用翻转了)
• 最后要判断字符串是否全0

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参考资料:

• LeetCode原题