LeetCode 1090. Largest Values From Labels

标签：use   xpl   color   problem   number   amp   代码   from   des   

Problem Description: We have a set of items: the i-th item has value values[i] and label labels[i].

Then, we choose a subset S of these items, such that:

• |S| <= num_wanted
• For every label L, the number of items in S with label L is <= use_limit.

Return the largest possible sum of the subset S.

Example 1:

Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], num_wanted = 3, use_limit = 1
Output: 9
Explanation: The subset chosen is the first, third, and fifth item.

Example 2:

Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], num_wanted = 3, use_limit = 2
Output: 12
Explanation: The subset chosen is the first, second, and third item.

今天周赛这道题没做出来，当时想到了才有贪心的思想，但是想得过于复杂，想对每一个label单独做一个优先队列，用一个map来映射label和对应的优先队列，另一个map来映射label和个数。
后来想到其实一个优先队列和一个记录label对应个数的map就可以解决。
代码如下：

public int largestValsFromLabels(int[] values, int[] labels, int num_wanted, int use_limit) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> (values[b] - values[a]));
        for(int i = 0; i < values.length; i++) {
            pq.offer(i);
        }
        Map<Integer, Integer> labelCounter = new HashMap<>();
        int res = 0;
        while(!pq.isEmpty() && num_wanted > 0) {
            int index = pq.poll();
            int value = values[index];
            int label = labels[index];
            
            if(labelCounter.getOrDefault(label, 0) < use_limit) {
                res += value;
                num_wanted--;
                labelCounter.put(label, labelCounter.getOrDefault(label, 0) + 1);
            } else {
                continue;
            }
        }
        return res;
    }

LeetCode 1090. Largest Values From Labels

标签：use   xpl   color   problem   number   amp   代码   from   des   

原文地址：https://www.cnblogs.com/rookielet/p/11037560.html