[LeetCode] 2341. Maximum Number of Pairs in Array

You are given a 0-indexed integer array nums. In one operation, you may do the following:

• Choose two integers in nums that are equal.
• Remove both integers from nums, forming a pair.

The operation is done on nums as many times as possible.

Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.

Example 1:

Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.

Example 2:

Input: nums = [1,1]
Output: [1,0]
Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.

Example 3:

Input: nums = [0]
Output: [0,1]
Explanation: No pairs can be formed, and there is 1 number leftover in nums.

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 100

数组能形成多少数对。

给你一个下标从 0 开始的整数数组 nums 。在一步操作中，你可以执行以下步骤：

从 nums 选出 两个 相等的 整数
从 nums 中移除这两个整数，形成一个 数对
请你在 nums 上多次执行此操作直到无法继续执行。

返回一个下标从 0 开始、长度为 2 的整数数组 answer 作为答案，其中 answer[0] 是形成的数对数目，answer[1] 是对 nums 尽可能执行上述操作后剩下的整数数目。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/maximum-number-of-pairs-in-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是 counting sort。需要遍历两次 input 数组，第一次遍历我们用一个哈希表把每个不同元素的出现次数记录下来；第二次遍历的时候，遍历哈希表的 keyset，对于每个不同的 key，如果他的出现次数 % 2 == 0，则他是一个 pair；否则他最后一定有一个落单的。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int[] numberOfPairs(int[] nums) {
        int pair = 0;
        int leftOver = 0;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }
        
        for (int num : map.keySet()) {
            int val = map.get(num);
            if (val % 2 == 1) {
                leftOver++;
            }
            pair += val / 2;
        }
        
        int[] res = new int[2];
        res[0] = pair;
        res[1] = leftOver;
        return res;
    }
}

 

LeetCode 题目总结