1342. Reduce Array Size to The Half

标签：col   integer   move   equal   array   eve   which   least   turn   

Given an array arr.  You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

Example 3:

Input: arr = [1,9]
Output: 1

Example 4:

Input: arr = [1000,1000,3,7]
Output: 1

Example 5:

Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5

Constraints:

• 1 <= arr.length <= 10^5
• arr.length is even.
• 1 <= arr[i] <= 10^5

class Solution {
    public int minSetSize(int[] arr) {
        Map<Integer, Integer> map = new HashMap();
        for(int i: arr){
            map.put(i, map.getOrDefault(i, 0) + 1);
        }
        int[] help = new int[map.values().size()];
        int siz = 0;
        List<Integer> list = new ArrayList(map.values());
        for(int i = 0; i < help.length; i++){
            help[i] = list.get(i);
            siz += help[i];
        }
        Arrays.sort(help);
        int res = 0, ind = help.length - 1;
        int le = siz / 2;
        while(siz > le && ind >= 0){
            res++;
            siz -= help[ind];
            ind--;
        }
        return res;
    }
}

记录出现的次数，然后减，注意最后是>不是>=
然后还要注意，Arrays.sort()不能用lambda表达式来实现int的排序，只能对Integer这种object使用

1342. Reduce Array Size to The Half

标签：col   integer   move   equal   array   eve   which   least   turn   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12254597.html