[LeetCode][C++]1768. Merge String Alternately

❓Question

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

• Example 1

  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

  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

  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.

💻Solution

class Solution {
public:
    string mergeAlternately(string word1, string word2) {
        // 단어 길이가 더 짧은 단어 기준으로 번갈아가면서 문자를 합침
        int minSize = word1.length() <= word2.length() ? word1.length() : word2.length();
        string longerStr = word1.length() <= word2.length() ? word2 : word1;

        string answer = "";
        answer.reserve(word1.length() + word2.length()); // 미리 메모리 확보

        for (int i = 0; i < minSize; i++)
        {
            answer.push_back(word1[i]); // 효율적으로 문자 추가
            answer.push_back(word2[i]);
        }  

        // 나머지 부분을 split해서 뒤에 붙임  
        if(longerStr.length() > minSize)  answer.append(longerStr.substr(minSize));

        return answer;
    }
};

• Complexity
  • Time: O(N+M)
  • Space: O(1)