LeetCode_ 2341_ How many pairs can an array form

Topic link

• https://leetcode.cn/problems/maximum-number-of-pairs-in-array/

Title Description

I'll give you a subscript from 0 The starting array of integers nums . In one step , You can do the following ：

• from nums elect Two equal Integers
• from nums Remove these two integers , To form a Number pair

Please come in nums Perform this operation several times on until it cannot be continued .

Returns a subscript from 0 Start 、 The length is 2 Array of integers for answer As the answer , among answer[0] Is the number of pairs formed ,answer[1] It's right nums Try to count the number of integers left after the above operation .

Example 1：

 Input ：nums = [1,3,2,1,3,2,2]
 Output ：[3,1]
 explain ：
nums[0]  and  nums[3]  Form a number pair , And from  nums  Remove ,nums = [3,2,3,2,2] .
nums[0]  and  nums[2]  Form a number pair , And from  nums  Remove ,nums = [2,2,2] .
nums[0]  and  nums[1]  Form a number pair , And from  nums  Remove ,nums = [2] .
 Cannot form more pairs . A total of  3  Pairs of numbers ,nums  The rest of the world is  1  A digital .

Example 2：

 Input ：nums = [1,1]
 Output ：[1,0]
 explain ：nums[0]  and  nums[1]  Form a number pair , And from  nums  Remove ,nums = [] .
 Cannot form more pairs . A total of  1  Pairs of numbers ,nums  The rest of the world is  0  A digital .

Example 3：

 Input ：nums = [0]
 Output ：[0,1]
 explain ： Cannot form a number of pairs ,nums  The rest of the world is  1  A digital .

Tips ：

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

Their thinking

Use one HashSet To traverse the array , For an element , If HashSet Contains this element , Then add one to the number of pairs , And then we can change this element from HashSet Remove , The last remaining number after several pairs of operations

AC Code

class Solution {
    
    public int[] numberOfPairs(int[] nums) {
    
        int num = 0;
        HashSet<Integer> hashSet = new HashSet<>();
        for (int i : nums) {
    
            if (hashSet.contains(i)) {
    
                hashSet.remove(i);
                num++;
            } else {
    
                hashSet.add(i);
            }
        }
        return new int[]{
    num, nums.length - 2 * num};
    }
}