交替合并字符串
难度:
标签:
题目描述
代码结果
运行时间: 16 ms, 内存: 16.7 MB
/*
* Problem Statement:
* Given two strings word1 and word2, merge them by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters to the merged string.
*
* Approach using Java Stream:
* 1. Use IntStream to iterate over the indices of both strings.
* 2. For each index, get the character from word1 and word2 if available.
* 3. Collect the characters to form the final merged string.
*/
import java.util.stream.Collectors;
import java.util.stream.IntStream;
public class Solution {
public String mergeAlternately(String word1, String word2) {
int maxLen = Math.max(word1.length(), word2.length());
return IntStream.range(0, maxLen)
.mapToObj(i -> {
StringBuilder sb = new StringBuilder();
if (i < word1.length()) sb.append(word1.charAt(i));
if (i < word2.length()) sb.append(word2.charAt(i));
return sb.toString();
})
.collect(Collectors.joining());
}
}
解释
方法:
The provided solution intends to merge two strings word1 and word2 by alternating characters from each. It begins by identifying the length of the shorter string (w). It then slices both strings into parts that have equivalent indices (c and d) and the leftover parts (e and f) when one string is longer than the other. In a loop, it alternates adding characters from c and d to a new string s, but it inserts them in reverse order, hence the need to reverse s at the end. After the loop, it appends any leftover characters from the longer string.
时间复杂度:
O(n)
空间复杂度:
O(n)
代码细节讲解
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为什么在合并字符时选择先将字符添加到结果字符串s的前面，然后在最后再进行一次字符串反转，而不是直接按顺序添加到s的后面？
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如果word1或word2为空字符串，这个解决方案是否还会正确运行，还是需要特殊处理？
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在实际应用中，是否有更简单或更直接的方法来实现这个字符串交替合并的功能而不需要额外的字符串反转操作？
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