Alternating merging strings of leetcode simple questions

subject

Here are two strings word1 and word2 . Please start from word1 Start , Merging strings by alternately adding letters . If one string is longer than another , Add the extra letters to the end of the merged string .
return Merged string .
Example 1：
Input ：word1 = “abc”, word2 = “pqr”
Output ：“apbqcr”
explain ： The string merge is shown below ：
word1： a b c
word2： p q r
After the merger ： a p b q c r
Example 2：
Input ：word1 = “ab”, word2 = “pqrs”
Output ：“apbqrs”
explain ： Be careful ,word2 Than word1 Long ,“rs” You need to append to the end of the merged string .
word1： a b
word2： p q r s
After the merger ： a p b q r s
Example 3：
Input ：word1 = “abcd”, word2 = “pq”
Output ：“apbqcd”
explain ： Be careful ,word1 Than word2 Long ,“cd” You need to append to the end of the merged string .
word1： a b c d
word2： p q
After the merger ： a p b q c d
Tips ：
1 <= word1.length, word2.length <= 100
word1 and word2 It's made up of lowercase letters
source ： Power button （LeetCode）

Their thinking

   Traverse two... Respectively word, Fill in the blank string in the order required by the title , Who grows whose tail is placed at the back .

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        n1=len(word1)
        n2=len(word2)
        word1=iter(word1)
        word2=iter(word2)
        temp=''
        if n1==n2:
            for i in range(n1):
                temp+=next(word1)
                temp+=next(word2)
        elif n1>n2:
            for i in range(n2):
                temp+=next(word1)
                temp+=next(word2)
            for i in range(n1-n2):
                temp+=next(word1)
        else:
            for i in range(n1):
                temp+=next(word1)
                temp+=next(word2)
            for i in range(n2-n1):
                temp+=next(word2)
        return temp