算法作业_23(2017.5.23第十四周)

575. Distribute Candies

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.    

暴力法： Time Limit Exceeded 

public class Solution {
    public int distributeCandies(int[] candies) {
        Arrays.sort(candies);
        int maxSort = 0;
        int maxKinds = 0;
        int count = 0;
        for(int i = 0 ;i<candies.length;i=i+count){
            maxSort++;
            count = 0;
            for(int j=i+1;j<candies.length;j++){
                if(candies[i]==candies[j]){
                    count++;
                }
            }
            if(count>maxKinds){
                maxKinds=count;
            }
        }
        return (maxSort>maxKinds/2)?maxSort:maxKinds/2;
        
    }
}

方法二：可以用HashSet将元素存一次，得到元素的种类，然后和总数的一半进行比较，如果种类的数目，多于总数字的半数，那么只能选择半数，否则不能平分。如果种类的数目少于总数字的半数，就可以选择种类的数目：

public class Solution {
    public int distributeCandies(int[] candies) {
        HashSet<Integer> set = new HashSet<Integer>();
        for(int num:candies){
            set.add(num);
        }
        return Math.min(set.size(),candies.length/2);
        
    }
}

class Solution {
public:
    int distributeCandies(vector<int>& candies) {
        unordered_map<int,int> Hash;
 
        for(int c :candies){
            Hash[c]++;
        }
        return min(Hash.size(),candies.size()/2);
    }
};