621. Task Scheduler

Description:

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:

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

简要题解：

采用贪心算法。

核心思想是： 每次挑选能执行的并且剩余最多的那种task来执行。没有这样的task时，就反复添加idle直到有可以执行的task。重复前面的步骤。

另外，要注意到，不需要详细地区分task的类型，只要能知道每种task剩下的数量即可。

代码：

class Solution {public:    int leastInterval(vector<char>& tasks, int n) {        vector<int> taskStats(26, 0);        int num = 0, i;        for (int i = 0; i < tasks.size(); i++)            taskStats[int(tasks[i]-'A')]++;        sort(taskStats.begin(), taskStats.end());        while (true) {            for (i = taskStats.size() - 1; i >= 0 && taskStats[i] != 0; i--) {                taskStats[i]--;                if (taskStats.size() - 1 - i == n) {                    i--;                    break;                }            }            sort(taskStats.begin(), taskStats.end());            if (taskStats.back() != 0) {                num += 1 + n;            } else {                num += taskStats.size() - 1 - i;                break;            }        }        return num;    }};