Last Updated on 2023-04-18 by Clay
題目
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 <= 100word1andword2consist of lowercase English letters.
題目會給我們兩個字串，而我們要做的就是將其交錯地排列成新的字串。比方說 abc 和 xyz，我們就要將其補成 a x b y c z，交錯地排列。
如果有一個字串長度較短，那麼新字串就直接補上還有剩餘的字串的尾部即可。
解題思路
這題由於是 Easy，也沒有太多巧妙地解法。我認為就將其交錯地組成新字串，接著判斷哪個字串較長，將其直接補上。
C++ 範例程式碼
class Solution {
public:
string mergeAlternately(string word1, string word2) {
// Init
int size1 = word1.size();
int size2 = word2.size();
std::string mergeWord;
// Reserve space for the merged word
mergeWord.reserve(size1 + size2);
// Merge
int i = 0, j = 0;
while (i < size1 && j < size2) {
mergeWord.push_back(word1[i++]);
mergeWord.push_back(word2[j++]);
}
// Append the remaining characters
if (i < size1) {
mergeWord.append(word1.substr(i));
}
else {
mergeWord.append(word2.substr(j));
}
return mergeWord;
}
};
Python 範例程式碼
class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
# Init
size1 = len(word1)
size2 = len(word2)
mergeWord = ""
# Merge
i, j = 0, 0
while i < size1 and j < size2:
mergeWord += word1[i]
mergeWord += word2[j]
i += 1
j += 1
# Append the remaining characters
if i < size1:
mergeWord += word1[i:]
else:
mergeWord += word2[j:]
return mergeWord