# [LeetCode 415. Add Strings](https://leetcode.com/explore/challenge/card/august-leetcoding-challenge-2021/614/week-2-august-8th-august-14th/3875/)
### Daily challenge Aug 9, 2021 (EASY)

>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.

:::info
**Example 1:**
**Input:** num1 = "11", num2 = "123"
**Output:** "134"
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:::info
**Example 2:**
**Input:** num1 = "456", num2 = "77"
**Output:** "533"
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:::info
**Example 3:**
**Input:** num1 = "0", num2 = "0"
**Output:** "0"
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:::warning
**Constraints:**
* 1 <= num1.length, num2.length <= 104
* num1 and num2 consist of only digits.
* num1 and num2 don't have any leading zeros except for the zero itself.
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---

### Approach 1 :
**`0 ms ( 100% )`** **`O()`**

* 讓 `num1` 保持是長度較長的字串，直接對 `num1` 進行運算。
* **`int i = num1.length() - 1`**、**`int j = num2.length() - 1`** 表示兩字串的個位數。
* 直接進行運算 **`sum = num1[i] + num2[j] - 96 + carry`**。
因為 `num2` 長度小於等於 `num1`，所以當 `num2` 遍歷完後還要考慮 `num1` 是否會有進位的問題。
---> `num1 = "9999"`、`num2 = "1"`。

class Solution {
public:
    string addStrings(string num1, string num2) {
        if(num1.length() < num2.length())   swap(num1, num2);
        
        string ans;
        int i = num1.length() - 1;
        int j = num2.length() - 1;
        int carry = 0;
        
        while(j >= 0){
            int sum = num1[i] + num2[j] - 96 + carry;
            carry = sum / 10;
            num1[i] = sum % 10 + 48;
            i--;
            j--;
        }
        
        while(i >= 0){
            int sum = num1[i] - 48 + carry;
            carry = sum / 10;
            num1[i] = sum % 10 + 48;
            i--;
        }
        if(carry == 1)  num1.insert(0, "1");
        
        return num1;
    }
};