[LeetCode] Distribute Candies 分糖果

Given an integer array with even length, where different numbers in this array represent different kinds of candies. Each number means one candy of the corresponding kind. You need to distribute these candies equally in number to brother and sister. Return the maximum number of kinds of candies the sister could gain.

Example 1:

Input: candies = [1,1,2,2,3,3]
Output: 3
Explanation:
There are three different kinds of candies (1, 2 and 3), and two candies for each kind.
Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too. 
The sister has three different kinds of candies. 

Example 2:

Input: candies = [1,1,2,3]
Output: 2
Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1]. 
The sister has two different kinds of candies, the brother has only one kind of candies. 

Note:

1. The length of the given array is in range [2, 10,000], and will be even.
2. The number in given array is in range [-100,000, 100,000].

这道题给我们一堆糖，每种糖的个数不定，分给两个人，让我们求其中一个人能拿到的最大的糖的种类数。那么我们想，如果总共有n个糖，平均分给两个人，每人得到n/2块糖，那么能拿到的最大的糖的种类数也就是n／2种，不可能再多，只可能再少。那么我们要做的就是统计出总共的糖的种类数，如果糖的种类数小于n/2，说明拿不到n/2种糖，最多能拿到的种类数数就是当前糖的总种类数，明白了这点就很容易了，我们利用集合set的自动去重复特性来求出糖的种类数，然后跟n/2比较，取二者之中的较小值返回即可，参加代码如下：

解法一：

class Solution {
public:
    int distributeCandies(vector<int>& candies) {
        unordered_set<int> s;
        for (int candy : candies) s.insert(candy);
        return min(s.size(), candies.size() / 2);
    }
};

下面这种方法叼的不行，直接用把上面的解法浓缩为了一行，有种显摆的感觉：

解法二：

class Solution {
public:
    int distributeCandies(vector<int>& candies) {
        return min(unordered_set<int>(candies.begin(), candies.end()).size(), candies.size() / 2);
    }
};

参考资料：

https://leetcode.com/problems/distribute-candies/

https://leetcode.com/problems/distribute-candies/discuss/102946/c-1-liner

https://leetcode.com/problems/distribute-candies/discuss/102939/java-8-one-line-solution-on

https://leetcode.com/problems/distribute-candies/discuss/102879/java-solution-3-lines-hashset

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