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:
The length of the given array is in range [2, 10,000], and will be even.
The number in given array is in range [-100,000, 100,000].
Solution：
思路：
如果總共有n個糖需频，平均分給兩個人修肠，每人得到n/2塊糖户盯，那么能拿到的最大的糖的種類數(shù)也就是n／2種嵌施，不可能再多莽鸭，只可能再少。那么我們要做的就是統(tǒng)計出總共的糖的種類數(shù)硫眨，如果糖的種類數(shù)小于n/2，說明拿不到n/2種糖礁阁，最多能拿到的種類數(shù)數(shù)就是當前糖的總種類數(shù)
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
public class Solution {
public int distributeCandies(int[] candies) {
Set<Integer> kinds = new HashSet<>();
for (int candy : candies) kinds.add(candy);
return kinds.size() >= candies.length / 2 ? candies.length / 2 : kinds.size();
}
}
