Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
- The length of both
num1andnum2is < 5100. - Both
num1andnum2contains only digits0-9. - Both
num1andnum2does not contain any leading zero. - You must not use any built-in BigInteger library or convert the inputs to integer directly.
两个只含有数字的字符串相加。
解法: 对于每个字符转成对应的整数，然后相加，结果在写入res。
Python:
class Solution(object):
def addStrings(self, num1, num2):
"""
:type num1: str
:type num2: str
:rtype: str
"""
result = []
i, j, carry = len(num1) - 1, len(num2) - 1, 0
while i >= 0 or j >= 0 or carry:
if i >= 0:
carry += ord(num1[i]) - ord(‘0‘);
i -= 1
if j >= 0:
carry += ord(num2[j]) - ord(‘0‘);
j -= 1
result.append(str(carry % 10))
carry /= 10
result.reverse()
return "".join(result)
C++:
class Solution {
public:
string addStrings(string num1, string num2) {
string res = "";
int m = num1.size(), n = num2.size(), i = m - 1, j = n - 1, carry = 0;
while (i >= 0 || j >= 0) {
int a = i >= 0 ? num1[i--] - ‘0‘ : 0;
int b = j >= 0 ? num2[j--] - ‘0‘ : 0;
int sum = a + b + carry;
res.insert(res.begin(), sum % 10 + ‘0‘);
carry = sum / 10;
}
return carry ? "1" + res : res;
}
};
