Problem

Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string.

You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.

Examples

Example 1:

Input:
num1 = "11", num2 = "123"
Output:
 "134"

Example 2:

Input:
num1 = "456", num2 = "77"
Output:
 "533"

Example 3:

Input:
num1 = "0", num2 = "0"
Output:
 "0"

Solution

Method 1 - Iterate from end of string and reverse

Code

C++
class Solution {
public:
    string addStrings(string num1, string num2) {
        int i = num1.size() - 1, j = num2.size() - 1;
        string ans;
        for (int c = 0; i >= 0 || j >= 0 || c; --i, --j) {
            int a = i < 0 ? 0 : num1[i] - '0';
            int b = j < 0 ? 0 : num2[j] - '0';
            c += a + b;
            ans += to_string(c % 10);
            c /= 10;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
Go
func addStrings(num1 string, num2 string) string {
	i, j := len(num1)-1, len(num2)-1
	ans := []byte{}
	for c := 0; i >= 0 || j >= 0 || c > 0; i, j = i-1, j-1 {
		if i >= 0 {
			c += int(num1[i] - '0')
		}
		if j >= 0 {
			c += int(num2[j] - '0')
		}
		ans = append(ans, byte(c%10+'0'))
		c /= 10
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return string(ans)
}
Java
class Solution {
    public String addStrings(String num1, String num2) {
        StringBuilder result = new StringBuilder();
        int i = num1.length() - 1;
        int j = num2.length() - 1;
        int carry = 0;


        while (i >= 0 || j >= 0) {
            int d1 = i >= 0 ? num1.charAt(i) - '0' : 0;
            int d2 = j >= 0 ? num2.charAt(j) - '0' : 0;

            int sum = (d1 + d2 + carry);
            
            carry = sum / 10;

            result.append(sum % 10);
            i--;
            j--;
        }

        if (carry == 1) {
            result.append("1");
        }

        return result.reverse().toString();
    }
}
JavaScript
/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var addStrings = function (num1, num2) {
    let i = num1.length - 1;
    let j = num2.length - 1;
    const ans = [];
    for (let c = 0; i >= 0 || j >= 0 || c; --i, --j) {
        c += i < 0 ? 0 : parseInt(num1.charAt(i), 10);
        c += j < 0 ? 0 : parseInt(num2.charAt(j), 10);
        ans.push(c % 10);
        c = Math.floor(c / 10);
    }
    return ans.reverse().join('');
};
Python
class Solution:
    def addStrings(self, num1: str, num2: str) -> str:
        i, j = len(num1) - 1, len(num2) - 1
        ans = []
        c = 0
        while i >= 0 or j >= 0 or c:
            a = 0 if i < 0 else int(num1[i])
            b = 0 if j < 0 else int(num2[j])
            c, v = divmod(a + b + c, 10)
            ans.append(str(v))
            i, j = i - 1, j - 1
        return "".join(ans[::-1])
Typescript
function addStrings(num1: string, num2: string): string {
    const res = [];
    let i = num1.length - 1;
    let j = num2.length - 1;
    let isOver = false;
    while (i >= 0 || j >= 0 || isOver) {
        const x = Number(num1[i--]) || 0;
        const y = Number(num2[j--]) || 0;
        const sum = x + y + (isOver ? 1 : 0);
        isOver = sum >= 10;
        res.push(sum % 10);
    }
    return res.reverse().join('');
}
Rust
impl Solution {
    pub fn add_strings(num1: String, num2: String) -> String {
        let mut res = vec![];
        let s1 = num1.as_bytes();
        let s2 = num2.as_bytes();
        let (mut i, mut j) = (s1.len(), s2.len());
        let mut is_over = false;
        while i != 0 || j != 0 || is_over {
            let mut sum = if is_over { 1 } else { 0 };
            if i != 0 {
                sum += (s1[i - 1] - b'0') as i32;
                i -= 1;
            }
            if j != 0 {
                sum += (s2[j - 1] - b'0') as i32;
                j -= 1;
            }
            is_over = sum >= 10;
            res.push((sum % 10).to_string());
        }
        res.into_iter().rev().collect()
    }
}

Complexity

  • Time: O(n)
  • Space: O(n)