621. Task Scheduler

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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:

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

解法1：

先找出出现频率最大的数字，假设频率为k。然后分开每一个数字使得两个数字之间间隔为n。
这样，会有k-1个间隔出现。所有频率小于k的数字都可以完全填充在这k-1个间隔里面。
那么，这个时候总长度就是：(k - 1) * (n + 1) + 频率为k的数字的个数。
（n+1)是因为每一个分段都是一个最高频的数字加上剩下的空， i.e. AXXXXXXXXX
C++

1

Java

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public class Solution {
    public int leastInterval(char[] tasks, int n) {
        int[] map = new int[26];
        for (char task: tasks) {
            map[task - 'A']++;
        }
        Arrays.sort(map);
        int maxFreq = map[25];
        int t = 0;
        for (int i = map.length - 1; i >= 0; i--) {
            if (map[i] == maxFreq) {
                t++;
            } else {
                break;
            }
        }
        return Math.max(tasks.length, (maxFreq - 1) * (n + 1) + t);
    }
}