问题描述
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
Note:
- The length of both num1 and num2 is < 110.
- Both num1 and num2 contains only digits 0-9.
- Both num1 and num2 does not contain any leading zero.
- You must not use any built-in BigInteger library or convert the inputs to integer directly.
思路分析
我的思路是完全模拟手算乘法，每次用乘数的一位乘以被乘数，与结果相加。这样做可以AC但是很慢（66 ms，只能beats 7.19%）。
参考了ChiangKaiShrek的解法 ，按位累加即可，不用每次都创建新的string来进行累加。
AC代码
class Solution {
public:
string multiply(string num1, string num2) {
int len1 = num1.size();
int len2 = num2.size();
string rst(len1+len2, '0');
for (int i = len1-1; i >= 0; --i){
int carry = 0;
for (int j = len2-1; j >= 0; --j){
int tmp = (rst[i+j+1] - '0') + (num1[i]-'0') * (num2[j]-'0') + carry;
rst[i+j+1] = char(tmp % 10 + '0');
carry = tmp / 10;
}
rst[i] += carry;
}
int i;
for(i = 0; i < len1+len2; ++i)
if (rst[i] != '0')
break;
if (i == len1+len2)
return "0";
return rst.substr(i);
}
};
Runtime: 6 ms, which beats 78.05% of cpp submissions.
得到”按位累加”的启发后改进了自己的思路，也取的了不错的效果。
string multiply(string num1, string num2) {
if (num1 == "0" || num2 == "0")
return "0";
int len1 = num1.size();
int len2 = num2.size();
string rst_s(len1+len2, '0');
vector<int> rst(len1+len2);
for (int i = len1 - 1; i >= 0; i--){
for (int j = len2 - 1; j >= 0; j--) {
rst[i+j+1] += (num1[i] - '0') * (num2[j] - '0');
}
}
int carry = 0;
int i = len1+len2-1;
while (i >= 0){
rst_s[i] = char(((carry + rst[i]) % 10) + '0');
carry = (carry + rst[i]) / 10;
--i;
}
i = 0;
while (i < len1+len2 && rst_s[i] == '0')
++i;
return rst_s.substr(i);
}
同样Runtime: 6 ms, which beats 78.05% of cpp submissions.
