Star

LeetCode Weekly Contest 621. Task Scheduler

Posted on | In |

Question

Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.

However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.

You need to return the least number of intervals the CPU will take to finish all the given tasks.

Example 1: 

1
2
3
Input: tasks = ['A','A','A','B','B','B'], n = 2
Output: 8
Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.

Note: The number of tasks is in the range [1, 10000]. The integer n is in the range [0, 100].

Explanation

这道题就是依次找到最max的frequency,之后顺次递减，如果还有剩余，就放回去。

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
public class Solution {
    public int leastInterval(char[] tasks, int n) {
        HashMap<Character, Integer> map = new HashMap<>();
        // store all tasks with its frequency
        for (char c : tasks) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        // store sorted frequency
        PriorityQueue<Map.Entry<Character, Integer>> queue = new PriorityQueue<>(
            (a,b) -> a.getValue() != b.getValue() ? b.getValue() - a.getValue() : a.getKey() - b.getKey());
        queue.addAll(map.entrySet());
        int count = 0;
        while(!queue.isEmpty()) {
            int k = n+1;
            List<Map.Entry> tmpList = new ArrayList<>();
            while (k > 0 && !queue.isEmpty()) {
                Map.Entry<Character, Integer> top = queue.poll();
                top.setValue(top.getValue()-1);
                tmpList.add(top);
                k --;
                count ++;
            }
            for (Map.Entry<Character, Integer> e:tmpList) {
                if (e.getValue() > 0)
                queue.add(e);
            }
            if (queue.isEmpty()) break;
            count = count + k;
        }
        return count;  
    }
}