Task Scheduler (M)
題目
You are given a char array representing tasks CPU need to do. It contains capital letters A to Z where each letter represents a different task. Tasks could be done without the original order of the array. Each task is done in one unit of time. For each unit of time, the CPU could complete either one task or just be idle.
However, there is a non-negative integer n that represents the cooldown period between two same tasks (the same letter in the array), that is that there must be at least n units of time between any two same tasks.
You need to return the least number of units of times that 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
There is at least 2 units of time between any two same tasks.
Example 2:
Input: tasks = ["A","A","A","B","B","B"], n = 0
Output: 6
Explanation: On this case any permutation of size 6 would work since n = 0.
["A","A","A","B","B","B"]
["A","B","A","B","A","B"]
["B","B","B","A","A","A"]
...
And so on.
Example 3:
Input: tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
Output: 16
Explanation:
One possible solution is
A -> B -> C -> A -> D -> E -> A -> F -> G -> A -> idle -> idle -> A -> idle -> idle -> A
Constraints:
- The number of tasks is in the range
[1, 10000]. - The integer
nis in the range[0, 100].
題意
給定一個字符陣列，將其進行排列，兩個相同的字符之間至少間隔n個字符，可以插入空字符，要求找到一個長度最短的符合條件的排列，
思路
貪心，先將字符按照出現次數降序排列，存入優先佇列，每次按序取出n+1個字符加入到排列中，如果不足n+1則用空字符替代，將這些取出字符的出現次數-1，如果不為0則再加入到優先佇列中，重復操作直到所有字符都被加入到排列中，
代碼實作
Java
class Solution {
public int leastInterval(char[] tasks, int n) {
int ans = 0;
int[] count = new int[26];
for (char c : tasks) {
count[c - 'A']++;
}
Queue<Character> q = new PriorityQueue<>((Character a, Character b) -> count[b - 'A'] - count[a - 'A']);
for (int i = 0; i < 26; i++) {
if (count[i] > 0) {
q.offer((char) ('A' + i));
}
}
while (!q.isEmpty()) {
Queue<Character> tmp = new LinkedList<>();
for (int i = 0; i < n + 1; i++) {
if (!q.isEmpty()) {
char top = q.poll();
count[top - 'A']--;
if (count[top - 'A'] != 0) {
tmp.offer(top);
}
ans++;
} else if (!tmp.isEmpty()) {
ans++;
}
}
while (!tmp.isEmpty()) {
q.offer(tmp.poll());
}
}
return ans;
}
}
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