621. Task Scheduler--任务调度

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 = 2
Output: 8
Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.

Note:

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

       这道题目的意思是：CPU在进行任务调度是，如果设置间隔为n，则相同的任务之间的间隔是n，如果所有的任务不够间隔n次出现，

就用idle表示，即让CPU空转。用大写字母从A-Z表示不同的任务，求出完成所有的任务调度的间隔数。

分析如下：首先统计每个任务出现的次数，因为任务的出现必须间隔n个，把CPU的每轮调度有n+1个，完整的调度有max-1轮，max为出现次数最多的

任务个数，然后剩下的最后一轮调度是前面max-1轮调度剩下的任务的个数，只需要统计出现次数最多的任务个数即可。代码中对数组a排序后，

max=a[25]，25-i为出现次数最多的任务个数。但是如果经过这些调度后仍然有漏掉的任务，可以用Math.max(tasks.length,(a[25]-1)*(n+1)+25-i))

解决这个问题。代码如下：

int [] a = new int[26];
    	for(char t:tasks){a[t-'A']++;}
    	Arrays.sort(a);
    	int i=25;
    	while(i>=0&&a[i]==a[25]){i--;}
    	return(Math.max(tasks.length, (a[25]-1)*(n+1)+25-i));