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:
Input: tasks = ["A","A","A","B","B","B"], n = 2 Output: 8 Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.
class Solution {
public:
/* 设置一个按次数排序的优先队列
* 每次从队列中前n+1 元素中每个抽取一个 次数--
* 思路2： 观察
* 取最大出现次数为主模块（有多个就视为一体）然后取出
* */
int leastInterval(vector<char>& tasks, int n) {
vector<int> times(26, 0); // 统计出现次数
for(char &c: tasks)
times[c-'A']++;
sort(times.begin(), times.end());
int maxs = times.back(), maxleft=0; // maxleft 记录相同最大值个数
for(int i=25;i>=0;i--){
if(times[i] == maxs) maxleft++;
}
// A--A--A--(Aleft)
return max(int(tasks.size()), (maxs-1)*(n+1)+maxleft);
}
};