621. Task Scheduler

621. Task Scheduler

• 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:

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

• 题目大意：给定一个char数组，代表CPU去完成的任务，给定一个数字n，代表任意两个相同的任务需要相隔n个时间间隔，求执行完所有任务所需的时间。

• 思路：

  1. 先统计词频，再排序，从后往前找到第一个不是最大词频的下标i，结果是tasks.length或(c[25] - 1) * (n + 1) + 25 – i中大的那一个，25-i就是最大词频的任务类，这个和我的思路是一样的。证明：最大词频是k，则创建k个块，每一块开头是最大词频的任务构成的(输入AACCCDDEEE，则开头是CE)，词频由大到小插入每一块。97.69%，10ms。
  2. 贪心，利用优先队列排序：队列中保存<类型，个数>的map，并且按照个数由大到小排序。按照词频由大到小取出n+1个或者队列中全部(若没有取出全部，则总长度要加上空闲个数)，再把词频-1之后不为0的放回队列中。直到队列空了为止。——其实也和前面相同，总是选择词频最大的填入每一块。

• 代码：

  class Solution {  public int leastInterval(char[] tasks, int n) {         int[] c = new int[26];      for (char t : tasks) {          c[t - 'A']++;      }      Arrays.sort(c);      int i = 25;      while (i >= 0 && c[i] == c[25]) {          i--;      }      return Math.max(tasks.length, (c[25] - 1) * (n + 1) + 25 - i);  }}

  第二种解法

  public class Solution {  public int leastInterval(char[] tasks, int n) {      Map<Character, Integer> counts = new HashMap<Character, Integer>();      for (char t : tasks) {          counts.put(t, counts.getOrDefault(t, 0) + 1);      }      PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a, b) -> b - a);      pq.addAll(counts.values());      int alltime = 0;      int cycle = n + 1;      while (!pq.isEmpty()) {          int worktime = 0;          List<Integer> tmp = new ArrayList<Integer>();          for (int i = 0; i < cycle; i++) {              if (!pq.isEmpty()) {                  tmp.add(pq.poll());                  worktime++;              }          }          for (int cnt : tmp) {              if (--cnt > 0) {                  pq.offer(cnt);              }          }          alltime += !pq.isEmpty() ? cycle : worktime;      }      return alltime;  }}