LeetCode 43. Multiply Strings

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.

 

Answer:

1.错位相加，只设一个carryBit的变量记录每次的进位，不用create extra space：

123 x 456 = 56088：

1:3x6 %= 8, c =1 --> 2: 2x6 + 5x3 +1 %= 8, c = 2 --> 3: 6x1+ 5x2 + 4x3 +2 %= 0, c =3 --> 4: 1x5 + 4 x2 + 3 %= 6, c = 1 --> 5: 4x1 + 1 = 5

public String multiply(String num1, String num2) {
        if(num1.length() == 0 || num2.length() ==0)
            return "";
        if(num1.equals("0") || num2.equals("0"))
            return "0";
        int carryBit = 0;
        String longStr = (num1.length() >= num2.length()) ? num1:num2;
        String shortStr = (num1.length() < num2.length()) ? num1:num2;
        String result = "";
        int i = 0;
        int j =0;
        while(i < longStr.length() && j < shortStr.length()){
            int sum = carryBit;
            int n = j;
            while(n < shortStr.length()){
                if(longStr.length()-1-i+n-j < longStr.length())
                    sum += (shortStr.charAt(shortStr.length()-1-n)- '0') * (longStr.charAt(longStr.length()-1-i+n-j)- '0');
                n++;
            }
            carryBit = sum /10;
            int digit = sum %10;
            result = digit + result;
            if(i == longStr.length()-1 )
                j++;
            if(i < longStr.length() -1)
                i++;
        }
        if(carryBit > 0){
            result = carryBit + result;
        }  
        return result;
    }

 

2. 输入的两个数和返回的数都是以字符串格式储存的，这样做的原因可能是这样可以计算超大数相乘，可以不受int或long的数值范围的约束，那么我们该如何来计算乘法呢，我们小时候都学过多位数的乘法过程，都是每位相乘然后错位相加，那么这里就是用到这种方法，参见网友JustDoIt的博客，把错位相加后的结果保存到一个一维数组中，然后分别每位上算进位，最后每个数字都变成一位，然后要做的是去除掉首位0，最后把每位上的数字按顺序保存到结果中即可。

c++:

class Solution {
public:
    string multiply(string num1, string num2) {
        string res;
        int n1 = num1.size(), n2 = num2.size();
        int k = n1 + n2 - 2, carry = 0;
        vector<int> v(n1 + n2, 0);
        for (int i = 0; i < n1; ++i) {
            for (int j = 0; j < n2; ++j) {
                v[k - i - j] += (num1[i] - '0') * (num2[j] - '0');
            }
        }
        for (int i = 0; i < n1 + n2; ++i) {
            v[i] += carry;
            carry = v[i] / 10;
            v[i] %= 10;
        }
        int i = n1 + n2 - 1;
        while (v[i] == 0) --i;
        if (i < 0) return "0";
        while (i >= 0) res.push_back(v[i--] + '0');
        return res;
    }
}