1. 题目
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.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.
2. 思路
模拟手算的过程,按位相乘并累加。
方便处理中间结果,每次保存的都是逆序的,然后反转。也可以每次都保持逆序结果,到最后的时候再反转过来。
3. 代码
耗时:26ms
class Solution {
public:
// 模拟乘法手算过程实现
string multiply(string num1, string num2) {
string sum;
sum.reserve(num1.length() * num2.length() + 2);
string prefix;
// 从低到高位逐个字符相乘,然后加入到总和中去。
for (int i = num2.length() - 1; i >= 0; i--) {
char c2 = num2[i];
string c_sum = multiply(num1, c2) + prefix;
sum = plus(sum, c_sum);
prefix += '0';
}
// 去掉前导0多余的0,至少保留一位
int i = 0;
while (sum[i] == '0' && i < sum.length() - 1) {i++;}
return sum.substr(i);
}
string multiply(string& n, char c) {
string r;
r.reserve(n.length() + 2);
int jingwei = 0;
for (int i = n.length() - 1; i>= 0; i--) {
int m = multiply(n[i], c);
m += jingwei;
r += m % 10 + '0';
jingwei = m / 10;
}
if (jingwei != 0) {
char ch = '0' + jingwei;
r += ch;
}
reverse(r.begin(), r.end());
return r;
}
int multiply(char c1, char c2) {
return (c1 - '0') * (c2 - '0');
}
string plus(string& n1, string& n2) {
string n3;
int l1 = n1.length();
int l2 = n2.length();
int min_v = min(l1, l2);
int max_v = max(l1, l2);
n3.reserve(max_v + 2);
char jingwei = '0';
for (int i = 1; i <= min_v; i++) {
char c1 = n1[l1 - i];
char c2 = n2[l2 - i];
int c_sum = plus(c1, c2, jingwei);
n3 += c_sum % 10 + '0';
jingwei = c_sum / 10 + '0';
}
string& big = n1;
if (l2 > l1) { big = n2; }
for (int i = min_v + 1; i <= max_v; i++) {
int c_sum = plus(big[max_v - i], jingwei);
jingwei = c_sum / 10 + '0';
n3 += c_sum % 10 + '0';
}
if (jingwei != '0') {
n3 += jingwei;
}
reverse(n3.begin(), n3.end());
return n3;
}
int plus(char c1, char c2) {
return c1 + c2 - 2 * '0';
}
int plus(char c1, char c2, char c3) {
return c1 + c2 + c3 - 3 * '0';
}
};
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