0621. Task Scheduler (M)

Task Scheduler (M)

题目

You are given a char array representing tasks CPU need to do. It contains capital letters A to Z where each letter represents a different task. Tasks could be done without the original order of the array. Each task is done in one unit of time. For each unit of time, the CPU could complete either one task or just be idle.

However, there is a non-negative integer n that represents the cooldown period between two same tasks (the same letter in the array), that is that there must be at least n units of time between any two same tasks.

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

Example 1:

Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: 
A -> B -> idle -> A -> B -> idle -> A -> B
There is at least 2 units of time between any two same tasks.

Example 2:

Input: tasks = ["A","A","A","B","B","B"], n = 0
Output: 6
Explanation: On this case any permutation of size 6 would work since n = 0.
["A","A","A","B","B","B"]
["A","B","A","B","A","B"]
["B","B","B","A","A","A"]
...
And so on.

Example 3:

Input: tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
Output: 16
Explanation: 
One possible solution is
A -> B -> C -> A -> D -> E -> A -> F -> G -> A -> idle -> idle -> A -> idle -> idle -> A

Constraints:

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

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题意

给定一个字符数组，将其进行排列，两个相同的字符之间至少间隔n个字符，可以插入空字符，要求找到一个长度最短的符合条件的排列。

思路

贪心。先将字符按照出现次数降序排列，存入优先队列。每次按序取出n+1个字符加入到排列中，如果不足n+1则用空字符替代，将这些取出字符的出现次数-1，如果不为0则再加入到优先队列中。重复操作直到所有字符都被加入到排列中。

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代码实现

Java

class Solution {
    public int leastInterval(char[] tasks, int n) {
        int ans = 0;
        int[] count = new int[26];
        for (char c : tasks) {
            count[c - 'A']++;
        }
        Queue<Character> q = new PriorityQueue<>((Character a, Character b) -> count[b - 'A'] - count[a - 'A']);
        for (int i = 0; i < 26; i++) {
            if (count[i] > 0) {
                q.offer((char) ('A' + i));
            }
        }
        while (!q.isEmpty()) {
            Queue<Character> tmp = new LinkedList<>();
            for (int i = 0; i < n + 1; i++) {
                if (!q.isEmpty()) {
                    char top = q.poll();
                    count[top - 'A']--;
                    if (count[top - 'A'] != 0) {
                        tmp.offer(top);
                    }
                    ans++;
                } else if (!tmp.isEmpty()) {
                    ans++;
                }
            }
            while (!tmp.isEmpty()) {
                q.offer(tmp.poll());
            }
        }
        return ans;
    }
}