LeetCode 75 - Merge Strings Alternately

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.

Example 1:

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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

Example 2:

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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

Example 3:

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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

Constraints:

1 <= word1.length, word2.length <= 100
word1 and word2 consist of lowercase English letters.

Solutions

For this problem, the first thing you can think of is to insert characters alternately. This is where you need to think about what to do with string length inconsistencies.

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class Solution {
public:
    string mergeAlternately(string word1, string word2) {
        int i{0};
        int j{0};
        string re;
        while (i < word1.size()) {
            re.push_back(word1[i++]);
            if (j < word2.size()) {
                re.push_back(word2[j++]);
            }
        }
        re += word2.substr(j);
        return re;
    }
};

Compared with the official solution, the above solution is less if judgment, but it also increases the difficulty of understanding. See the official solution:

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class Solution {
public:
    string mergeAlternately(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        int i = 0, j = 0;
        string ans;
        
        while (i < m || j < n) {
            if (i < m) {
                ans.push_back(word1[i]);
                ++i;
            }
            if (j < n) {
                ans.push_back(word2[j]);
                ++j;
            }
        }
        return ans;
    }
};