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Formatted question description: https://leetcode.ca/all/1768.html
1768. Merge Strings Alternately
Level
Easy
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
andword2
consist of lowercase English letters.
Solution
Merge the two strings as long as both strings have remaining letters. When a string does not have remaining letters, append the remaining letters of the other string to the merged string.
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class Solution { public String mergeAlternately(String word1, String word2) { StringBuffer sb = new StringBuffer(); int length1 = word1.length(), length2 = word2.length(); int index1 = 0, index2 = 0; while (index1 < length1 && index2 < length2) { sb.append(word1.charAt(index1++)); sb.append(word2.charAt(index2++)); } while (index1 < length1) sb.append(word1.charAt(index1++)); while (index2 < length2) sb.append(word2.charAt(index2++)); return sb.toString(); } } ############ 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(); } }
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// OJ: https://leetcode.com/problems/merge-strings-alternately/ // Time: O(M + N) // Space: O(1) class Solution { public: string mergeAlternately(string a, string b) { int i = 0, j = 0, M = a.size(), N = b.size(); string ans; while (i < M || j < N) { if (i < M) ans += a[i++]; if (j < N) ans += b[j++]; } return ans; } };
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class Solution: def mergeAlternately(self, word1: str, word2: str) -> str: return ''.join(a + b for a, b in zip_longest(word1, word2, fillvalue='')) ############ class Solution: def mergeAlternately(self, word1: str, word2: str) -> str: cur1 = 0 cur2 = 0 res = "" while cur1 < len(word1) and cur2 < len(word2): res += word1[cur1] + word2[cur2] cur1 += 1 cur2 += 1 if cur1 != len(word1): res += word1[cur1:] if cur2 != len(word2): res += word2[cur2:] return res
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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) }
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function mergeAlternately(word1: string, word2: string): string { const res = []; const n = Math.max(word1.length, word2.length); for (let i = 0; i < n; i++) { word1[i] && res.push(word1[i]); word2[i] && res.push(word2[i]); } return res.join(''); }
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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() } }