[LeetCode] 43. Multiply Strings 字符串相乘

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.

求两个字符串数字的相乘，输入的两个数和返回的数都是以字符串格式储存，这样做的原因可能是计算超大数相乘不受数值范围的约束。题目也要求不能用内置长整数计算。

解法：用数组来记录每一位的计算结果，最后在转成字符串就可以了。

Java:

public String multiply(String num1, String num2) {
    int m = num1.length(), n = num2.length();
    int[] pos = new int[m + n];
   
    for(int i = m - 1; i >= 0; i--) {
        for(int j = n - 1; j >= 0; j--) {
            int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0'); 
            int p1 = i + j, p2 = i + j + 1;
            int sum = mul + pos[p2];

            pos[p1] += sum / 10;
            pos[p2] = (sum) % 10;
        }
    }  
    
    StringBuilder sb = new StringBuilder();
    for(int p : pos) if(!(sb.length() == 0 && p == 0)) sb.append(p);
    return sb.length() == 0 ? "0" : sb.toString();
}　　

Python:

class Solution(object):
    def multiply(self, num1, num2):
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        num1, num2 = num1[::-1], num2[::-1]
        res = [0] * (len(num1) + len(num2))
        for i in xrange(len(num1)):
            for j in xrange(len(num2)):
                res[i + j] += int(num1[i]) * int(num2[j])
                res[i + j + 1] += res[i + j] / 10
                res[i + j] %= 10

        # Skip leading 0s.
        i = len(res) - 1
        while i > 0 and res[i] == 0:
            i -= 1

        return ''.join(map(str, res[i::-1]))

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;
    }
};

　　

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