621. Task Scheduler

原题：

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 twosame 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:

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

思考过程&解题思路：一开始没什么好的思路，借鉴了别人的思路，受了一点启发，最后自己完成。先把出现次数最多的字母排好，其它字母往里插入。最后有两种情况：插入之后仍有空缺，则返回值是(n + 1) * (k - 1) + j。其中k是出现次数最多的字母的出现次数，j是出现次数为k的字母个数；都插满之后仍然有字母剩余，那么剩余字母总可以无缝插进去，这时候返回值是tasks.length。

结果代码：

public int leastInterval(char[] tasks, int n) {        int[] alphabet = new int[26];//int[n]的值表示第n + 1个字母在tasks里出现的次数        int len = tasks.length;        for (int i = 0;i < len;i++)            alphabet[tasks[i] - 'A']++;        Arrays.sort(alphabet);        int i = 1,max = alphabet[25];//max是出现次数最大值        while (alphabet[25 - i] == alphabet[25]) i++;//看看有几个最大值，这会影响结果        return Math.max((n + 1) * (max - 1) + i,len);    }