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1768. 交替合并字符串


1768. 交替合并字符串

🟢   🔖  双指针 字符串  🔗 力扣open in new window LeetCodeopen in new window

题目

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 <= 100
  • word1 and word2 consist of lowercase English letters.

题目大意

给你两个字符串 word1word2 。请你从 word1 开始,通过交替添加字母来合并字符串。如果一个字符串比另一个字符串长,就将多出来的字母追加到合并后字符串的末尾。

返回 合并后的字符串

示例 1:

输入: word1 = "abc", word2 = "pqr"

输出: "apbqcr"

解释: 字符串合并情况如下所示:

word1: a b c

word2: p q r

合并后: a p b q c r

示例 2:

输入: word1 = "ab", word2 = "pqrs"

输出: "apbqrs"

解释: 注意,word2 比 word1 长,"rs" 需要追加到合并后字符串的末尾。

word1: a b

word2: p q r s

合并后: a p b q r s

示例 3:

输入: word1 = "abcd", word2 = "pq"

输出: "apbqcd"

解释: 注意,word1 比 word2 长,"cd" 需要追加到合并后字符串的末尾。

word1: a b c d

word2: p q

合并后: a p b q c d

提示:

  • 1 <= word1.length, word2.length <= 100
  • word1word2 由小写英文字母组成

解题思路

  1. 初始化指针

    • 使用两个指针分别遍历 word1word2,从头到尾交替添加字符。
  2. 交替合并

    • 在一个循环中,依次从 word1word2 中取出字符,直到其中一个字符串的字符用完。
  3. 处理剩余字符

    • 如果 word1word2 中还有剩余字符,将其直接追加到结果字符串的末尾。

复杂度分析

  • 时间复杂度O(n + m),其中 nm 分别是 word1word2 的长度,需要遍历每个字符串一次。
  • 空间复杂度O(n + m),用于存储合并后的字符串。

代码

/**
 * @param {string} word1
 * @param {string} word2
 * @return {string}
 */
var mergeAlternately = function (word1, word2) {
	const m = word1.length,
		n = word2.length;
	let i = 0,
		res = '';
	while (i < m && i < n) {
		res += word1[i] + word2[i];
		i++;
	}
	res += m > n ? word1.slice(i, m) : word2.slice(i, n);
	return res;
};

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