[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);
}
}
posted @   CNoodle  阅读(29)  评论(0编辑  收藏  举报
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