415 Add Strings 字符串相加
Description:
Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
The length of both num1 and num2 is < 5100.
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.
题目描述:
给定两个字符串形式的非负整数 num1 和num2 ,计算它们的和。
注意:
num1 和num2 的长度都小于 5100.
num1 和num2 都只包含数字 0-9.
num1 和num2 都不包含任何前导零。
你不能使用任何內建 BigInteger 库, 也不能直接将输入的字符串转换为整数形式。
思路:
参考LeetCode #67 Add Binary 二进制求和
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
string addStrings(string num1, string num2)
{
string result = "";
int carry = 0, i = num1.size() - 1, j = num2.size() - 1;
while (i >= 0 or j >= 0 or carry != 0)
{
if (i >= 0) carry += num1[i--] - '0';
if (j >= 0) carry += num2[j--] - '0';
result += to_string(carry % 10);
carry /= 10;
}
reverse(result.begin(), result.end());
return result;
}
};
Java:
class Solution {
public String addStrings(String num1, String num2) {
StringBuilder sb = new StringBuilder();
int carry = 0, i = num1.length() - 1, j = num2.length() - 1;
while (i >= 0 || j >= 0 || carry != 0) {
if (i >= 0) carry += num1.charAt(i--) - '0';
if (j >= 0) carry += num2.charAt(j--) - '0';
sb.append(carry % 10);
carry /= 10;
}
return sb.reverse().toString();
}
}
Python:
class Solution:
def addStrings(self, num1: str, num2: str) -> str:
num1, num2, max_length, result, carry = num1[::-1], num2[::-1], max(len(num1), len(num2)), '', 0
for i in range(max_length):
first, second = int(num1[i]) if i < len(num1) else 0, int(num2[i]) if i < len(num2) else 0
result += str((first + second + carry) % 10)
carry = int((first + second + carry) / 10)
if carry > 0:
result += str(carry)
return result[::-1]