[LeetCode] 621. Task Scheduler 任务调度
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].
一个表示CPU要做的任务数组。用大写字母A到Z来表示，不同的字母表示不同的任务。任务完成顺序不限，每个任务可在一个时间内完成，对于每个时间，CPU可以完成一个任务或空闲。但在两个相同任务之间要有一个冷却时间间隔n，间隔内CPU可以执行其他不同的任务或是处于空闲状态。求CPU完成所有给定任务所需的最小时间。
解法：先用哈希表或者数组对任务出现的次数进行统计，找出任务出现次数的最大值max_count(有可能是多个)，然后把这个任务(或者任务对)之间按n个间隔进行分隔排列，然后把其它任务插到这些间隔里面。由于其它任务出现的次数少于max_count所以肯定能放下，如有空位就是需要空闲(idle)。最后，还要注意的是，如果空位都插满之后还有任务，那就随便在这些间隔里面插入就可以，因为间隔长度肯定会大于n，在这种情况下就是任务的总数是最小所需时间。
情况1：AAABBCD，n=2。分隔为：AXXAXXA -> ABCABDA（其中，X表示需要填充任务或者idle的间隔）
情况2：AAAABBBBCCDE，n=3。分隔为：ABXXABXXABXXAB -> ABCDABCEABXXAB 或者 ABCXABCXABDEAB, etc. 此时要加上出现次数也是max_count的任务的个数same_max_count(相当于末尾多了这几个)。
情况3：AAABBCDEF，n=2。分隔为：AXXAXXA -> ABCEABDFA 或者 ABCEABDAF, etc. 此时就是task的长度是最小时间。
最少的任务时间公式：max(len(tasks), (max_count-1) * (n+1) + other_max_count)
Java: O(1) time O(1) space
public class Solution {
public int leastInterval(char[] tasks, int n) {
int[] counter = new int[26];
int max = 0;
int maxCount = 0;
for(char task : tasks) {
counter[task - 'A']++;
if(max == counter[task - 'A']) {
maxCount++;
}
else if(max < counter[task - 'A']) {
max = counter[task - 'A'];
maxCount = 1;
}
}
int partCount = max - 1;
int partLength = n - (maxCount - 1);
int emptySlots = partCount * partLength;
int availableTasks = tasks.length - max * maxCount;
int idles = Math.max(0, emptySlots - availableTasks);
return tasks.length + idles;
}
}
Java: O(N) time O(26) space
// (c[25] - 1) * (n + 1) + 25 - i is frame size
// when inserting chars, the frame might be "burst", then tasks.length takes precedence
// when 25 - i > n, the frame is already full at construction, the following is still valid.
public class Solution {
public int leastInterval(char[] tasks, int n) {
int[] c = new int[26];
for(char t : tasks){
c[t - 'A']++;
}
Arrays.sort(c);
int i = 25;
while(i >= 0 && c[i] == c[25]) i--;
return Math.max(tasks.length, (c[25] - 1) * (n + 1) + 25 - i);
}
}
Python: O(1) time O(1) space
class Solution(object):
def leastInterval(self, tasks, n):
"""
:type tasks: List[str]
:type n: int
:rtype: int
"""
count = collections.defaultdict(int)
max_count = 0
for task in tasks:
count[task] += 1
max_count = max(max_count, count[task])
result = (max_count-1) * (n+1)
for count in count.values():
if count == max_count:
result += 1
return max(result, len(tasks))
C++:
class Solution {
public:
int leastInterval(vector<char>& tasks, int n) {
int mx = 0, mxCnt = 0;
vector<int> cnt(26, 0);
for (char task : tasks) {
++cnt[task - 'A'];
if (mx == cnt[task - 'A']) {
++mxCnt;
} else if (mx < cnt[task - 'A']) {
mx = cnt[task - 'A'];
mxCnt = 1;
}
}
int partCnt = mx - 1;
int partLen = n - (mxCnt - 1);
int emptySlots = partCnt * partLen;
int taskLeft = tasks.size() - mx * mxCnt;
int idles = max(0, emptySlots - taskLeft);
return tasks.size() + idles;
}
};
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