simulation/multiply-strings
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2023-12-01
Multiply Strings
描述
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.
分析
高精度乘法。
代码
模拟乘法
// Multiply Strings
// Time Complexity: O(n*m), Space Complexity: O(n+m)
class Solution {
public String multiply(String num1, String num2) {
int m = num1.length(), n = num2.length();
int[] z = 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 sum = mul + z[i+j+1];
z[i + j + 1] = sum % 10;
z[i + j] += sum / 10; // carry
}
}
StringBuilder sb = new StringBuilder();
for(int x : z) {
if(!(sb.length() == 0 && x == 0)) // skip prefix zeros
sb.append(x);
}
return sb.length() == 0 ? "0" : sb.toString();
}
}
// TODO
转化成整数数组，一个字符对应一个整数
常见的做法是把每个字符转化为一个 int，一一对应，形成一个 int 数组。
// Multiply Strings
// 一个字符对应一个int
// 时间复杂度O(n*m)，空间复杂度O(n+m)
public class Solution {
public String multiply(String num1, String num2) {
BigInt bigInt1 = new BigInt(num1);
BigInt bigInt2 = new BigInt(num2);
BigInt result = BigInt.multiply(bigInt1, bigInt2);
return result.toString();
}
// 一个字符对应一个int
static class BigInt {
private final int[] d;
public BigInt(String s) {
this.d = fromString(s);
}
public BigInt(int[] d) {
this.d = d;
}
private static int[] fromString(String s) {
int[] d = new int[s.length()];
for (int i = s.length() - 1, j = 0; i >= 0; --i)
d[j++] = Character.getNumericValue(s.charAt(i));
return d;
}
@Override
public String toString() {
final StringBuilder sb = new StringBuilder();
for (int i = d.length - 1; i >= 0; --i) {
sb.append(Character.forDigit(d[i], 10));
}
return sb.toString();
}
public static BigInt multiply(BigInt x, BigInt y) {
int[] z = new int[x.d.length + y.d.length];
for (int i = 0; i < x.d.length; ++i) {
for (int j = 0; j < y.d.length; ++j) {
z[i + j] += x.d[i] * y.d[j];
z[i + j + 1] += z[i + j] / 10;
z[i + j] %= 10;
}
}
// find the first 0 from right to left
int i = z.length - 1;
for (; i > 0 && z[i] == 0; --i) /* empty */;
if (i == z.length - 1) {
return new BigInt(z);
} else { // make a copy
int[] tmp = new int[i + 1];
System.arraycopy(z, 0, tmp, 0, i + 1);
return new BigInt(tmp);
}
}
}
}
// Multiply Strings
// @author 连城 (http://weibo.com/lianchengzju)
// 一个字符对应一个int
// 时间复杂度O(n*m)，空间复杂度O(n+m)
typedef vector<int> bigint;
bigint make_bigint(string const& repr) {
bigint n;
transform(repr.rbegin(), repr.rend(), back_inserter(n),
[](char c) { return c - '0'; });
return n;
}
string to_string(bigint const& n) {
string str;
transform(find_if(n.rbegin(), prev(n.rend()),
[](char c) { return c > '\0'; }), n.rend(), back_inserter(str),
[](char c) { return c + '0'; });
return str;
}
bigint operator*(bigint const& x, bigint const& y) {
bigint z(x.size() + y.size());
for (size_t i = 0; i < x.size(); ++i)
for (size_t j = 0; j < y.size(); ++j) {
z[i + j] += x[i] * y[j];
z[i + j + 1] += z[i + j] / 10;
z[i + j] %= 10;
}
return z;
}
class Solution {
public:
string multiply(string num1, string num2) {
return to_string(make_bigint(num1) * make_bigint(num2));
}
};
转化成整数数组，9 个字符对应一个 64 位整数
一个字符用一个 int 表示，其实是比较浪费内存空间的，因为一个 int64 的最大值是$2^{63}-1=9223372036854775807$，可以与 19 个字符对应，由于有乘法，减半，则至少可以与 9 个字符一一对应。
// Multiply Strings
// 9个字符对应一个 long
// 时间复杂度O(n*m)，空间复杂度O(n+m)
public class Solution {
public String multiply(String num1, String num2) {
BigInt bigInt1 = BigInt.fromString(num1);
BigInt bigInt2 = BigInt.fromString(num2);
BigInt result = BigInt.multiply(bigInt1, bigInt2);
return result.toString();
}
// 9个字符对应一个 long
static class BigInt {
/** 一个数组元素对应9个十进制位，即数组是亿进制的
* 因为 1000000000 * 1000000000 没有超过 2^63-1
*/
final static int BIGINT_RADIX = 1000000000;
final static int RADIX_LEN = 9;
/** 万进制整数. */
private final long[] digits;
public BigInt(long[] digits) {
this.digits = digits;
}
private static BigInt fromString(String s) {
long[] digits;
if (s.length() % RADIX_LEN == 0) {
digits = new long[s.length() / RADIX_LEN];
} else {
digits = new long[s.length() / RADIX_LEN + 1];
}
for (int i = s.length(), k = 0; i > 0; i -= RADIX_LEN) {
long tmp = 0;
for (int j = Math.max(0, i - RADIX_LEN); j < i; ++j) {
tmp = tmp * 10 + Character.getNumericValue(s.charAt(j));
}
digits[k++] = tmp;
}
return new BigInt(digits);
}
@Override
public String toString() {
final StringBuilder sb = new StringBuilder(
Long.toString(digits[digits.length-1]));
for (int i = digits.length - 2; i >= 0; --i) {
sb.append(String.format("%0" + RADIX_LEN + "d", digits[i]));
}
return sb.toString();
}
public static BigInt multiply(BigInt x, BigInt y) {
long[] z = new long[x.digits.length + y.digits.length];
for (int i = 0; i < x.digits.length; ++i) {
for (int j = 0; j < y.digits.length; ++j) {
z[i + j] += x.digits[i] * y.digits[j];
z[i + j + 1] += z[i + j] / BIGINT_RADIX;
z[i + j] %= BIGINT_RADIX;
}
}
// find the first 0 from right to left
int i = z.length - 1;
for (; i > 0 && z[i] == 0; --i) /* empty */;
if (i == z.length - 1) {
return new BigInt(z);
} else { // make a copy
long[] tmp = new long[i + 1];
System.arraycopy(z, 0, tmp, 0, i + 1);
return new BigInt(tmp);
}
}
}
}
// Multiply Strings
// 9个字符对应一个int64_t
// 时间复杂度O(n*m/81)，空间复杂度O((n+m)/9)
/** 大整数类. */
class BigInt {
public:
/**
* @brief 构造函数，将字符串转化为大整数.
* @param[in] s 输入的字符串
* @return 无
*/
BigInt(string s) {
vector<int64_t> result;
result.reserve(s.size() / RADIX_LEN + 1);
for (int i = s.size(); i > 0; i -= RADIX_LEN) { // [i-RADIX_LEN, i)
int temp = 0;
const int low = max(i - RADIX_LEN, 0);
for (int j = low; j < i; j++) {
temp = temp * 10 + s[j] - '0';
}
result.push_back(temp);
}
elems = result;
}
/**
* @brief 将整数转化为字符串.
* @return 字符串
*/
string toString() {
stringstream result;
bool started = false; // 用于跳过前导0
for (auto i = elems.rbegin(); i != elems.rend(); i++) {
if (started) { // 如果多余的0已经都跳过，则输出
result << setw(RADIX_LEN) << setfill('0') << *i;
} else {
result << *i;
started = true; // 碰到第一个非0的值，就说明多余的0已经都跳过
}
}
if (!started) return "0"; // 当x全为0时
else return result.str();
}
/**
* @brief 大整数乘法.
* @param[in] x x
* @param[in] y y
* @return 大整数
*/
static BigInt multiply(const BigInt &x, const BigInt &y) {
vector<int64_t> z(x.elems.size() + y.elems.size(), 0);
for (size_t i = 0; i < y.elems.size(); i++) {
for (size_t j = 0; j < x.elems.size(); j++) { // 用y[i]去乘以x的各位
// 两数第i, j位相乘，累加到结果的第i+j位
z[i + j] += y.elems[i] * x.elems[j];
if (z[i + j] >= BIGINT_RADIX) { // 看是否要进位
z[i + j + 1] += z[i + j] / BIGINT_RADIX; // 进位
z[i + j] %= BIGINT_RADIX;
}
}
}
while (z.back() == 0) z.pop_back(); // 没有进位，去掉最高位的0
return BigInt(z);
}
private:
typedef long long int64_t;
/** 一个数组元素对应9个十进制位，即数组是亿进制的
* 因为 1000000000 * 1000000000 没有超过 2^63-1
*/
const static int BIGINT_RADIX = 1000000000;
const static int RADIX_LEN = 9;
/** 万进制整数. */
vector<int64_t> elems;
BigInt(const vector<int64_t> num) : elems(num) {}
};
class Solution {
public:
string multiply(string num1, string num2) {
BigInt x(num1);
BigInt y(num2);
return BigInt::multiply(x, y).toString();
}
};