算法作業_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將元素存一次，得到元素的種類，然後和總數的一半進行比較，如果種類的數目，多於總數字的半數，那麼只能選擇半數，否則不能平分。如果種類的數目少於總數字的半數，就可以選擇種類的數目：

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);
    }
};