题目
给你一个用字符数组 tasks 表示的 CPU 需要执行的任务列表，用字母 A 到 Z 表示，以及一个冷却时间 n。每个周期或时间间隔允许完成一项任务。任务可以按任何顺序完成，但有一个限制：两个 相同种类 的任务之间必须有长度为 n 的冷却时间。
返回完成所有任务所需要的 最短时间间隔 。
示例 1：
输入：tasks = ["A","A","A","B","B","B"], n = 2
输出：8
解释：A -> B -> (待命) -> A -> B -> (待命) -> A -> B
在本示例中，两个相同类型任务之间必须间隔长度为 n = 2 的冷却时间，而执行一个任务只需要一个单位时间，所以中间出现了（待命）状态。
示例 2：
输入：tasks = ["A","A","A","B","B","B"], n = 0
输出：6
解释：在这种情况下，任何大小为 6 的排列都可以满足要求，因为 n = 0
["A","A","A","B","B","B"]
["A","B","A","B","A","B"]
["B","B","B","A","A","A"]
...
诸如此类
示例 3：
输入：tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
输出：16
解释：一种可能的解决方案是：
A -> B -> C -> A -> D -> E -> A -> F -> G -> A -> (待命) -> (待命) -> A -> (待命) -> (待命) -> A
代码实现
class Solution {
public int leastInterval(char[] tasks, int n) {
Map<Character, Integer> freq = new HashMap<Character, Integer>();
for (char ch : tasks) {
freq.put(ch, freq.getOrDefault(ch, 0) + 1);
}
// 任务种类数
int m = freq.size();
List<Integer> nextValid = new ArrayList<Integer>();
List<Integer> rest = new ArrayList<Integer>();
Set<Map.Entry<Character, Integer>> entrySet = freq.entrySet();
for (Map.Entry<Character, Integer> entry : entrySet) {
int value = entry.getValue();
nextValid.add(1);
rest.add(value);
}
int time = 0;
for (int i = 0; i < tasks.length; ++i) {
++time;
int minNextValid = Integer.MAX_VALUE;
for (int j = 0; j < m; ++j) {
if (rest.get(j) != 0) {
minNextValid = Math.min(minNextValid, nextValid.get(j));
}
}
time = Math.max(time, minNextValid);
int best = -1;
for (int j = 0; j < m; ++j) {
if (rest.get(j) != 0 && nextValid.get(j) <= time) {
if (best == -1 || rest.get(j) > rest.get(best)) {
best = j;
}
}
}
nextValid.set(best, time + n + 1);
rest.set(best, rest.get(best) - 1);
}
return time;
}
}
