Multiply Strings
Problem
Given two non-negative integers num1 and num2 represented as strings, return their product as a string. Do not convert directly to BigInteger or call any built-in big-number library.
num1 = "123", num2 = "456""56088"def multiply(num1, num2):
if num1 == "0" or num2 == "0": return "0"
m, n = len(num1), len(num2)
res = [0] * (m + n)
for i in range(m - 1, -1, -1):
for j in range(n - 1, -1, -1):
mul = (ord(num1[i]) - 48) * (ord(num2[j]) - 48)
p1, p2 = i + j, i + j + 1
total = mul + res[p2]
res[p2] = total % 10
res[p1] += total // 10
return "".join(map(str, res)).lstrip("0") or "0"function multiply(num1, num2) {
if (num1 === "0" || num2 === "0") return "0";
const m = num1.length, n = num2.length;
const res = new Array(m + n).fill(0);
for (let i = m - 1; i >= 0; i--) {
for (let j = n - 1; j >= 0; j--) {
const mul = (num1.charCodeAt(i) - 48) * (num2.charCodeAt(j) - 48);
const p1 = i + j, p2 = i + j + 1;
const total = mul + res[p2];
res[p2] = total % 10;
res[p1] += Math.floor(total / 10);
}
}
return res.join("").replace(/^0+/, "") || "0";
}class Solution {
public String multiply(String num1, String num2) {
if (num1.equals("0") || num2.equals("0")) return "0";
int m = num1.length(), n = num2.length();
int[] res = 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 total = mul + res[p2];
res[p2] = total % 10;
res[p1] += total / 10;
}
}
StringBuilder sb = new StringBuilder();
for (int d : res) if (!(sb.length() == 0 && d == 0)) sb.append(d);
return sb.length() == 0 ? "0" : sb.toString();
}
}string multiply(string num1, string num2) {
if (num1 == "0" || num2 == "0") return "0";
int m = num1.size(), n = num2.size();
vector<int> res(m + n, 0);
for (int i = m - 1; i >= 0; i--) {
for (int j = n - 1; j >= 0; j--) {
int mul = (num1[i] - '0') * (num2[j] - '0');
int p1 = i + j, p2 = i + j + 1;
int total = mul + res[p2];
res[p2] = total % 10;
res[p1] += total / 10;
}
}
string out;
for (int d : res) if (!(out.empty() && d == 0)) out += (char)('0' + d);
return out.empty() ? "0" : out;
}Explanation
We multiply two number-strings the way you would do long multiplication by hand, but with one clean trick: figure out exactly which output slot each digit-pair lands in.
The product of an m-digit number and an n-digit number has at most m + n digits, so we set up a result array res of that size, all zeros.
The key fact: multiplying digit i of num1 by digit j of num2 always affects positions i + j and i + j + 1 in res. We add the product into the lower slot p2 = i + j + 1, keep the units digit there with % 10, and carry the tens into the higher slot p1 = i + j.
After processing every pair, res holds the digits in order. We join them into a string and strip any leading zeros (falling back to "0" if everything cancels).
Example: "123" * "456". Each pair of digits is multiplied and dropped into its i+j slots with carries, and the array reads out as "56088".