Tcdian / keep

今天不想做,所以才去做。
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621. Task Scheduler #275

Open Tcdian opened 3 years ago

Tcdian commented 3 years ago

621. Task Scheduler

给定一个用字符数组表示的 CPU 需要执行的任务列表。其中包含使用大写的 A - Z 字母表示的26 种不同种类的任务。任务可以以任意顺序执行,并且每个任务都可以在 1 个单位时间内执行完。CPU 在任何一个单位时间内都可以执行一个任务,或者在待命状态。

然而,两个 相同种类 的任务之间必须有长度为 n 的冷却时间,因此至少有连续 n 个单位时间内 CPU 在执行不同的任务,或者在待命状态。

你需要计算完成所有任务所需要的 最短时间

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

Note

Tcdian commented 3 years ago

Solution

/**
 * @param {character[]} tasks
 * @param {number} n
 * @return {number}
 */
var leastInterval = function(tasks, n) {
    const cache = new Array(26).fill(0);
    for (let i = 0; i < tasks.length; i++) {
        cache[tasks[i].codePointAt(0) - 65] += 1;
    }
    cache.sort((a, b) => b - a);
    const bucket = cache[0];
    let taskEqualBucket = 0;
    while(taskEqualBucket < 26 && cache[taskEqualBucket] === bucket) {
        taskEqualBucket++;
    }
    return Math.max((bucket - 1) * (n + 1) + taskEqualBucket, tasks.length);
};
function leastInterval(tasks: string[], n: number): number {
    const cache: number[] = new Array(26).fill(0);
    for (let i = 0; i < tasks.length; i++) {
        cache[tasks[i].codePointAt(0) as number - 65] += 1;
    }
    cache.sort((a, b) => b - a);
    const bucket = cache[0];
    let taskEqualBucket = 0;
    while(taskEqualBucket < 26 && cache[taskEqualBucket] === bucket) {
        taskEqualBucket++;
    }
    return Math.max((bucket - 1) * (n + 1) + taskEqualBucket, tasks.length);
};