README_EN.md

1768. Merge Strings Alternately

中文文档

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:

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.

Solutions

Python3

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

Java

class Solution {
    public String mergeAlternately(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        StringBuilder res = new StringBuilder();
        for (int i = 0; i < m || i < n; ++i) {
            if (i < m) {
                res.append(word1.charAt(i));
            }
            if (i < n) {
                res.append(word2.charAt(i));
            }
        }
        return res.toString();
    }
}

C++

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

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