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.

Note: The number of tasks is in the range [1, 10000]. The integer n is in the range [0, 100].

思路： 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。 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。

class Solution { public int leastInterval(char[] tasks, int n) { int[] freq = new int[26]; int maxFreq = 0, maxFreqCount = 0; for(int i = 0; i < tasks.length; i++){ freq[tasks[i] - 'A']++; } for(int i = 0; i < 26; i++){ if(freq[i] > maxFreq){ maxFreq = freq[i]; maxFreqCount = 1; } else if(freq[i] == maxFreq){ maxFreqCount++; } } return Math.max(tasks.length, (maxFreq - 1) * (n + 1) + maxFreqCount); }}

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