leetcode 621. Task Scheduler

最新推荐文章于 2025-11-23 22:29:35 发布

原创最新推荐文章于 2025-11-23 22:29:35 发布 · 148 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#maxTask

leetcode 专栏收录该内容

14 篇文章

订阅专栏

本文深入探讨了CPU任务调度算法，特别是在存在冷却间隔约束的情况下，如何优化任务执行顺序以减少总执行时间。通过实例分析，介绍了如何计算在给定任务集和冷却间隔下，CPU完成所有任务所需的最短时间。

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.

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

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        int ans = 0;
        int freq[26] = {0};
        int theMaxCount = 0;
        int theFreqCount = 0;
        for(int i = 0; i < tasks.size(); i++){
            freq[tasks[i] - 'A'] ++;
            if(freq[tasks[i] - 'A'] > theMaxCount){
                theMaxCount = freq[tasks[i] - 'A'];
                theFreqCount = 1;
            }else if(freq[tasks[i] - 'A'] == theMaxCount){
                theFreqCount++;
            }
        }
        ans = (theMaxCount - 1) * (n + 1) + theFreqCount;
        ans = max(ans, int(tasks.size()));
        return ans;
    }
};