/**
Given a characters array tasks, representing the tasks a CPU needs to do, where
each letter represents a different task. Tasks could be done in any order. 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.
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:
1 <= task.length <= 10⁴
tasks[i] is upper-case English letter.
The integer n is in the range [0, 100].
Related Topics贪心 | 数组 | 哈希表 | 计数 | 排序 | 堆（优先队列）
👍 1050, 👎 0
*/
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
class Task {
char taskCh;
int number;
int lastTime;
}
public int leastInterval(char[] tasks, int n) {
Map<Character, Task> taskMap = new HashMap<>();
PriorityQueue<Task> pq = new PriorityQueue<>(new Comparator<Task>() {
@Override
public int compare(Task t1, Task t2) {
return t2.number - t1.number;
}
});
for (char task : tasks) {
if (!taskMap.containsKey(task)) {
Task newTask = new Task();
newTask.taskCh = task;
newTask.number = 0;
newTask.lastTime = -1;
taskMap.put(task, newTask);
}
Task taskObj = taskMap.get(task);
taskObj.number = taskObj.number + 1;
}
for (Task taskObj : taskMap.values()) {
pq.add(taskObj);
}
int result = 0;
while (!pq.isEmpty()) {
List<Task> taskList = new ArrayList<>();
while (!pq.isEmpty()) {
Task curTask = pq.poll();
if (curTask.lastTime >= 0 && (result - curTask.lastTime) <= n) {
taskList.add(curTask);
} else {
curTask.lastTime = result;
curTask.number = curTask.number - 1;
if (curTask.number > 0) {
taskList.add(curTask);
}
break;
}
}
if (taskList.size() > 0) {
for (Task curTask : taskList) {
pq.add(curTask);
}
}
result++;
}
return result;
}
}
//leetcode submit region end(Prohibit modification and deletion)
