给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。
注意：不能使用任何内置的 BigInteger 库或直接将输入转换为整数。
示例 1:
输入: num1 = "2", num2 = "3" 输出: "6"
示例 2:
输入: num1 = "123", num2 = "456" 输出: "56088"
提示：
1 <= num1.length, num2.length <= 200num1和num2只能由数字组成。num1和num2都不包含任何前导零，除了数字0本身。
class Solution:
def multiply(self, num1: str, num2: str) -> str:
def mul(b, i):
j, t = m - 1, 0
while j >= 0 or t:
if j >= 0:
a = int(num1[j])
t += a * b
res[i] += t % 10
if res[i] >= 10:
res[i] %= 10
res[i + 1] += 1
i, j = i + 1, j - 1
t //= 10
m, n = len(num1), len(num2)
res = [0] * (m + n)
for i in range(n):
b = int(num2[n - 1 - i])
mul(b, i)
while len(res) > 1 and res[-1] == 0:
res.pop()
return ''.join([str(v) for v in res[::-1]])class Solution {
public String multiply(String num1, String num2) {
int m = num1.length(), n = num2.length();
int[] res = new int[m + n];
for (int i = 0; i < n; ++i) {
int b = num2.charAt(n - 1 - i) - '0';
mul(num1, b, i, res);
}
StringBuilder ans = new StringBuilder();
for (int v : res) {
ans.append(v);
}
while (ans.length() > 1 && ans.charAt(ans.length() - 1) == '0') {
ans.deleteCharAt(ans.length() - 1);
}
return ans.reverse().toString();
}
private void mul(String A, int b, int i, int[] res) {
for (int j = A.length() - 1, t = 0; j >= 0 || t > 0; --j) {
if (j >= 0) {
int a = A.charAt(j) - '0';
t += a * b;
}
res[i++] += t % 10;
if (res[i - 1] >= 10) {
res[i - 1] %= 10;
++res[i];
}
t /= 10;
}
}
}class Solution {
public:
string multiply(string num1, string num2) {
int m = num1.size(), n = num2.size();
vector<int> res(m + n);
for (int i = 0; i < n; ++i) {
int b = num2[n - 1 - i] - '0';
mul(num1, b, i, res);
}
string ans = "";
for (int v : res) ans.push_back(v + '0');
while (ans.size() > 1 && ans.back() == '0') ans.pop_back();
reverse(ans.begin(), ans.end());
return ans;
}
void mul(string A, int b, int i, vector<int>& res) {
for (int j = A.size() - 1, t = 0; j >= 0 || t > 0; --j) {
if (j >= 0) {
int a = A[j] - '0';
t += a * b;
}
res[i++] += t % 10;
if (res[i - 1] >= 10) {
res[i - 1] %= 10;
++res[i];
}
t /= 10;
}
}
};func multiply(num1 string, num2 string) string {
m, n := len(num1), len(num2)
res := make([]int, m+n)
mul := func(b, i int) {
for j, t := m-1, 0; j >= 0 || t > 0; i, j = i+1, j-1 {
if j >= 0 {
a := int(num1[j] - '0')
t += a * b
}
res[i] += t % 10
if res[i] >= 10 {
res[i] %= 10
res[i+1]++
}
t /= 10
}
}
for i := 0; i < n; i++ {
b := num2[n-1-i] - '0'
mul(int(b), i)
}
var ans []byte
for _, v := range res {
ans = append(ans, byte(v+'0'))
}
for len(ans) > 1 && ans[len(ans)-1] == '0' {
ans = ans[:len(ans)-1]
}
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
ans[i], ans[j] = ans[j], ans[i]
}
return string(ans)
}function multiply(num1: string, num2: string): string {
if ([num1, num2].includes('0')) return '0';
const n1 = num1.length,
n2 = num2.length;
let ans = '';
for (let i = 0; i < n1; i++) {
let cur1 = parseInt(num1.charAt(n1 - i - 1), 10);
let sum = '';
for (let j = 0; j < n2; j++) {
let cur2 = parseInt(num2.charAt(n2 - j - 1), 10);
sum = addString(sum, cur1 * cur2 + '0'.repeat(j));
}
ans = addString(ans, sum + '0'.repeat(i));
}
return ans;
}
function addString(s1: string, s2: string): string {
const n1 = s1.length,
n2 = s2.length;
let ans = [];
let sum = 0;
for (let i = 0; i < n1 || i < n2 || sum > 0; i++) {
let num1 = i < n1 ? parseInt(s1.charAt(n1 - i - 1), 10) : 0;
let num2 = i < n2 ? parseInt(s2.charAt(n2 - i - 1), 10) : 0;
sum += num1 + num2;
ans.unshift(sum % 10);
sum = Math.floor(sum / 10);
}
return ans.join('');
}function multiply(num1: string, num2: string): string {
if (num1 === '0' || num2 === '0') {
return '0';
}
const n = num1.length;
const m = num2.length;
const res = [];
for (let i = 0; i < n; i++) {
const a = Number(num1[n - i - 1]);
let sum = 0;
for (let j = 0; j < m || sum !== 0; j++) {
const b = Number(num2[m - j - 1] ?? 0);
sum += a * b + (res[i + j] ?? 0);
res[i + j] = sum % 10;
sum = Math.floor(sum / 10);
}
}
return res.reverse().join('');
}impl Solution {
pub fn multiply(num1: String, num2: String) -> String {
if num1 == "0" || num2 == "0" {
return String::from("0");
}
let (num1, num2) = (num1.as_bytes(), num2.as_bytes());
let (n, m) = (num1.len(), num2.len());
let mut res = vec![];
for i in 0..n {
let a = num1[n - i - 1] - b'0';
let mut sum = 0;
let mut j = 0;
while j < m || sum != 0 {
if i + j == res.len() {
res.push(0)
}
let b = num2.get(m - j - 1).unwrap_or(&b'0') - b'0';
sum += a * b + res[i + j];
res[i + j] = sum % 10;
sum /= 10;
j += 1;
}
}
res.into_iter()
.rev()
.map(|v| char::from(v + b'0'))
.collect()
}
}