[LeetCode] 43. Multiply Strings 字符串相乘 All LeetCode Questions List 题目汇总

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

给定两个以字符表示的数字，求两个数字乘积，数字可以是任意大非负。以字符形式可以计算超大数相乘，不受int或long的数值范围的约束。

方法：如果给的数字不超出范围，可以直接转为整数相乘。但如果很大，就不可以直接相乘，而是第二个数的每一位和第一个数相乘然后错位相加，处理好进位和最后的数字。

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++:

// Time:  O(m * n)
// Space: O(m + n)

class Solution {
public:
    string multiply(string num1, string num2) {
        const auto char_to_int = [](const char c) { return c - '0'; };
        const auto int_to_char = [](const int i) { return i + '0'; };

        vector<int> n1;
        transform(num1.rbegin(), num1.rend(), back_inserter(n1), char_to_int);
        vector<int> n2;
        transform(num2.rbegin(), num2.rend(), back_inserter(n2), char_to_int);

        vector<int> tmp(n1.size() + n2.size());
        for(int i = 0; i < n1.size(); ++i) {
            for(int j = 0; j < n2.size(); ++j) {
                tmp[i + j] += n1[i] * n2[j];
                tmp[i + j + 1] += tmp[i + j] / 10;
                tmp[i + j] %= 10;
            }
        }
            
        string res;
        transform(find_if(tmp.rbegin(), prev(tmp.rend()),
                         [](const int i) { return i != 0; }),
                  tmp.rend(), back_inserter(res), int_to_char);
        return res;
    }
};

　　

　　

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