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:
<li><code>1 <= word1.length, word2.length <= 100</code></li>
<li><code>word1</code> and <code>word2</code> consist of lowercase English letters.</li>
Solutions
Solution 1: Direct Simulation
We traverse the two strings word1 and word2, take out the characters one by one, and append them to the result string. The Python code can be simplified into one line.
The time complexity is O(m + n), where m and n are the lengths of the two strings respectively. Ignoring the space consumption of the answer, the space complexity is O(1).
PythonJavaC++GoTypeScriptRustC
class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
return ''.join(a + b for a, b in zip_longest(word1, word2, fillvalue=''))(code-box)
class Solution {
public String mergeAlternately(String word1, String word2) {
int m = word1.length(), n = word2.length();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < m || i < n; ++i) {
if (i < m) {
ans.append(word1.charAt(i));
}
if (i < n) {
ans.append(word2.charAt(i));
}
}
return ans.toString();
}
}(code-box)
class Solution {
public:
string mergeAlternately(string word1, string word2) {
int m = word1.size(), n = word2.size();
string ans;
for (int i = 0; i < m || i < n; ++i) {
if (i < m) ans += word1[i];
if (i < n) ans += word2[i];
}
return ans;
}
};(code-box)
func mergeAlternately(word1 string, word2 string) string {
m, n := len(word1), len(word2)
ans := make([]byte, 0, m+n)
for i := 0; i < m || i < n; i++ {
if i < m {
ans = append(ans, word1[i])
}
if i < n {
ans = append(ans, word2[i])
}
}
return string(ans)
}(code-box)
function mergeAlternately(word1: string, word2: string): string {
const ans: string[] = [];
const [m, n] = [word1.length, word2.length];
for (let i = 0; i < m || i < n; ++i) {
if (i < m) {
ans.push(word1[i]);
}
if (i < n) {
ans.push(word2[i]);
}
}
return ans.join('');
}(code-box)
impl Solution {
pub fn merge_alternately(word1: String, word2: String) -> String {
let s1 = word1.as_bytes();
let s2 = word2.as_bytes();
let n = s1.len().max(s2.len());
let mut res = vec![];
for i in 0..n {
if s1.get(i).is_some() {
res.push(s1[i]);
}
if s2.get(i).is_some() {
res.push(s2[i]);
}
}
String::from_utf8(res).unwrap()
}
}(code-box)
char* mergeAlternately(char* word1, char* word2) {
int m = strlen(word1);
int n = strlen(word2);
char* ans = malloc(sizeof(char) * (n + m + 1));
int i = 0;
int j = 0;
while (i + j != m + n) {
if (i < m) {
ans[i + j] = word1[i];
i++;
}
if (j < n) {
ans[i + j] = word2[j];
j++;
}
}
ans[n + m] = '\0';
return ans;
}(code-box)
