leetcode-mid-others-621. Task Scheduler

mycode 53.01%

这个题在纸上画一画就知道啦，只要出现次数最多的字母能够满足要求，其他更少的字母穿插在其中，间隔就更满足<n啦，当然，最后不要忘记加上尾巴哦，尾巴和出现次数最多的字母的种类有关哦！

class Solution(object):

    def leastInterval(self, tasks, n):

        """

        :type tasks: List[str]

        :type n: int

        :rtype: int

        """

        from collections import Counter

        count = Counter(tasks)

        total = 1

        maxcount = count.most_common()[0][1]

        print(maxcount)

        for num,c in count.most_common()[1:]:

            if c == maxcount:

                total += 1

        return max(len(tasks),(maxcount-1)*(n+1)+total)

参考：

class Solution(object):

    def leastInterval(self, tasks, n):

        """

        :type tasks: List[str]

        :type n: int

        :rtype: int

        """

        lenT = len(tasks)

        res = rem = max_cnt = 0

        dic = collections.defaultdict(int)

        for ch in tasks:

            dic[ch]+=1

        for val in dic.values():

            if val>max_cnt:

                max_cnt = val

                rem = 0

            elif val==max_cnt:

                rem+=1 #与max_cnt出现频率相同的其他元素个数

        tmp = (max_cnt-1)*(n+1)+1 + rem # n很长的情况

        res = max(lenT, tmp)

        return res

巴特西

leetcode-mid-others-621. Task Scheduler

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