Distribute Candies–LeetCode#575

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.

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.

 思路:一个整形数组平均分成两份，求分出来的不同值有多少。就是求数组中整数不同的有几种与平均分后每组分到多少个的比较，返回比较下来小的值。

代码如下：

import java.util.Arrays;
/**
 * Created by Poldi on 2017/7/7.
 */
public class DistributeCandies {
    public int distributeCandies(int[] candies) {
        int num = candies.length/2;
        int different = 1;
        Arrays.sort(candies);
        for (int i=0;i<candies.length-1;i++){
            if (candies[i]!=candies[i+1]){
                different++;
            }
        }
        System.out.println(different);
        if (different>num){
            return num;
        }else if (different<num){
            return different;
        }else {
            return num;
        }
    }
    public static void main(String[] args) {
        DistributeCandies distributeCandies=new DistributeCandies();
        int[] ints={100000,0,100000,0,100000,0,100000,0,100000,0,100000,0};
        System.out.println(distributeCandies.distributeCandies(ints));
    }
}

另外有种更简便的思路，使用Set能直接把原数组中重复的整数去除，Set的实现类的集合对象中不能够有重复元素，HashSet也一样他是使用了一种标识来确定元素的不重复，HashSet用一种算法来保证HashSet中的元素是不重复的。从而可获取不同整数的个数，与平均分后的数量比较。
代码如下：

public int distributeCandies(int[] candies) {
    int num = candies.length/2;
    Set<Integer> candy = new HashSet<>();
    for (int i=0;i<candies.length;i++){
        candy.add(candies[i]);
    }
    if (candy.size()>num){
        return num;
    }else if (candy.size()<num){
        return candy.size();
    }else {
        return num;
    }
}