[LeetCode] 2744. Find Maximum Number of String Pairs

You are given a 0-indexed array words consisting of distinct strings.

The string words[i] can be paired with the string words[j] if:

The string words[i] is equal to the reversed string of words[j].
0 <= i < j < words.length.
Return the maximum number of pairs that can be formed from the array words.

Note that each string can belong in at most one pair.

Example 1:
Input: words = [“cd”,”ac”,”dc”,”ca”,”zz”]
Output: 2
Explanation: In this example, we can form 2 pair of strings in the following way:

• We pair the 0th string with the 2nd string, as the reversed string of word[0] is “dc” and is equal to words[2].
• We pair the 1st string with the 3rd string, as the reversed string of word[1] is “ca” and is equal to words[3].
  It can be proven that 2 is the maximum number of pairs that can be formed.

Example 2:
Input: words = [“ab”,”ba”,”cc”]
Output: 1
Explanation: In this example, we can form 1 pair of strings in the following way:

• We pair the 0th string with the 1st string, as the reversed string of words[1] is “ab” and is equal to words[0].
  It can be proven that 1 is the maximum number of pairs that can be formed.

Example 3:
Input: words = [“aa”,”ab”]
Output: 0
Explanation: In this example, we are unable to form any pair of strings.

Constraints:
1 <= words.length <= 50
words[i].length == 2
words consists of distinct strings.
words[i] contains only lowercase English letters.

最大字符串配对数目。

给你一个下标从 0 开始的数组 words ，数组中包含 互不相同 的字符串。
如果字符串 words[i] 与字符串 words[j] 满足以下条件，我们称它们可以匹配：
字符串 words[i] 等于 words[j] 的反转字符串。
0 <= i < j < words.length
请你返回数组 words 中的 最大 匹配数目。
注意，每个字符串最多匹配一次。

思路

题目给的是一些互不相同的字符串，所以一开始我用一个 hashset 把所有字符串存起来。因为题目要我们找的匹配字符串其实是找两个互为回文的字符串，所以这里我写一个 reverse 函数判断回文，对于某个单词，我得到他的回文之后去 hashset 判断这个回文是否存在于 hashset。注意这里有一个 corner case 是如果是类似 aaa 或者 zz 这种本身就是回文的单词，我们需要忽略。

复杂度

时间O(n)
空间O(n)

代码

Java实现

class Solution {
	public int maximumNumberOfStringPairs(String[] words) {
		Set<String> set = new HashSet<>();
		for (String word : words) {
			if (!word.equals(reverse(word))) {
				set.add(word);
			}
		}

		int count = 0;
		for (String word : words) {
			if (set.contains(word) && set.contains(reverse(word))) {
				count++;
				set.remove(word);
			}
		}
		return count;
	}

	private String reverse(String s) {
		char[] letters = s.toCharArray();
		int left = 0;
		int right = s.length() - 1;
		while (left < right) {
			char temp = letters[left];
			letters[left] = letters[right];
			letters[right] = temp;
			left++;
			right--;
		}
		return String.valueOf(letters);
	}
}