575. Distribute Candies

题目描述【Leetcode】

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].

这道题就是把数组平均分为两组，其中一组包含种类最多，求该种类个数

方法一：
就是先排好序，然后就找不同的数字即可

class Solution {
public:
    int distributeCandies(vector<int>& candies) {
        int n = candies.size();
        if(n == 0) return 0;
       sort(candies.begin(),candies.end());
       int count = 1;
       int temp = candies[0];
        for(int i = 1; i < n; i++ ){
            if(candies[i] != candies[i-1] && count < n/2)count++;
        }
        return count;
    }
};

方法二：删除重复项之后大小与之前数组作比较：

class Solution {
public:
    int distributeCandies(vector<int>& candies) {
        int n = candies.size();
        if(n == 0) return 0;
        sort(candies.begin(),candies.end());
        candies.erase(unique(candies.begin(),candies.end()),candies.end());
        int t = candies.size();
        if(t >= n/2) return n/2;
        if(t  < n/2) return t;
    }
};

方法三：用unordered_set，类似于方法二，但是运行更快：

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