## Description
>You are given two strings `word1` and `word2`. Merge the strings by adding letters in alternating order, starting with `word1`. If a string is longer than the other, append the additional letters onto the end of the merged string.
>
>Return the merged string.

## Constraints:
> `1 <= word1.length, word2.length <= 100`
> `word1` and `word2` consist of lowercase English letters.

## Example1
Input: word1 = "abc", word2 = "pqr"
Output: "apbqcr"
Explanation: The merged string will be merged as so:
word1: a b c
word2: p q r
merged: a p b q c r

## Example2
Input: word1 = "ab", word2 = "pqrs"
Output: "apbqrs"
Explanation: Notice that as word2 is longer, "rs" is appended to the end.
word1: a b
word2: p q r s
merged: a p b q r s
## Example3
Input: word1 = "abcd", word2 = "pq"
Output: "apbqcd"
Explanation: Notice that as word1 is longer, "cd" is appended to the end.
word1: a b c d
word2: p q
merged: a p b q c d

## Solution
當下直覺的作法，用i來計數，寫while迴圈判斷i有沒有大於兩字串長度，如果沒有那就依序把字串傳入ans裡，回傳ans得到答案。

* Time complexity: O(n)
* Space complexity: O(n)

## Code
### C++

class Solution {
public:
    string mergeAlternately(string word1, string word2) {
        string ans = "";
        int i = 0;
        while(i < word1.length() || i < word2.length())
        {
            if(i < word1.length())
            {
                ans += word1[i];
            }    
            if(i < word2.length())
            {
                ans += word2[i];
            }
            count++; 
        }    
        return ans;
    }
};

### Python3

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        ans = []
        i = 0
        while i < len(word1) or i < len(word2):
            if i < len(word1):
                ans.append(word1[i])
            if i < len(word2):
                ans.append(word2[i])
            i = i+1
        return ''.join(ans)