LeetCode43 Multiply Strings 字串相乘
阿新 • • 發佈:2019-02-04
題目描述:
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.
Example 1:
Input: num1 = "2", num2 = "3"
Output: "6"
Example 2:
Input: num1 = "123", num2 = "456"
Output: "56088"
Note:
The length of both num1 and num2 is < 110.
Both num1 and num2 contain only digits 0-9.
Both num1 and num2 do not contain any leading zero, except the number 0 itself.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
題源: here;完整實現:here
思路:
不用想多了,就是小學的乘法規則就可以了:從個位開始乘。當然,每一次的結果都是存在字串中,也就是說我們還需要完成字串的加法工作,實現的程式碼如下:
class Solution {
public:
string sum2string(string num1, string num2){
if (num1.size() == 0) return num2;
string result;
int carryBit = 0;
while(num1.size() && num2.size ()){
int S = num1[0] - '0' + num2[0] - '0' + carryBit;
result.push_back(S % 10 + '0');
carryBit = S / 10;
num1.erase(0, 1); num2.erase(0, 1);
}
if (num1.size()){
while (num1.size()){
int S = num1[0] - '0' + carryBit;
result.push_back(S % 10 + '0');
carryBit = S / 10;
num1.erase(0, 1);
}
}
else{
while (num2.size()){
int S = num2[0] - '0' + carryBit;
result.push_back(S % 10 + '0');
carryBit = S / 10;
num2.erase(0, 1);
}
}
if (carryBit > 0) result.push_back(carryBit + '0');
return result;
}
string multiply(string num1, string num2) {
if ((num1.size() == 1 && num1[0] == '0') || (num2.size() == 1 && num2[0] == '0')) return "0";
string result;
for (int i = num2.size()-1; i >= 0; i--){
string bitM;
for (int j = 0; j < num2.size() - i - 1; j++) bitM.push_back('0');
int carryBit = 0, digit2 = num2[i] - '0';
for (int j = num1.size() - 1; j >= 0; j--){
int digit1 = num1[j] - '0';
int M = digit1*digit2 + carryBit;
bitM.push_back(M % 10+'0');
carryBit = M / 10;
}
if (carryBit > 0) bitM.push_back(carryBit + '0');
result = sum2string(result, bitM);
}
reverse(result.begin(), result.end());
return result;
}
};