621. Task Scheduler

这又是一道没有固定套路，只是纯看自己能不能想出来解法的题目。
https://leetcode.com/problems/task-scheduler/discuss/104500/Java-O(n)-time-O(1)-space-1-pass-no-sorting-solution-with-detailed-explanation

class Solution {
public:
    int leastInterval(vector<char>& tasks, int n) {
        vector<int> cnts(26, 0);
        int maxCnt = 0;
        int maxCntNum = 0;
        for (char c : tasks) {
            cnts[c-'A']++;
            if (cnts[c-'A'] == maxCnt) 
                ++maxCntNum;
            else if (cnts[c-'A'] > maxCnt) {
                maxCnt = cnts[c-'A'];
                maxCntNum = 1;
            }
        }
        int partCount = maxCnt-1;
        int partLen = n - (maxCntNum - 1);
        int emptySlots = partCount * partLen;
        int availableTasks = tasks.size() - maxCnt * maxCntNum;
        int idles = max(0, emptySlots - availableTasks);
        return tasks.size() + idles;
    }
};

总的来说就是贪心，把出现频率最高的安排好了，其他的放到它之间就行了。
详细的看链接