Problem Description: We have a set of items: the i -th item has value values[i] and label labels[i] . Then, we choosea subset S of these items, such that: |S| = num_wanted For every label L , the number of items in S withlabel L is = use_li
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 inS
with labelL
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], = 3, use_limit = 1 Output: 9 Explanation: The subset chosen is the first, third, and fifth item.
num_wanted
Example 2:
Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], = 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:num_wanted
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; }