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Saturday, July 22, 2017

[Leetcode] Task Scheduler, Solution

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:
  1. The number of tasks is in the range [1, 10000].
  2. The integer n is in the range [0, 100].


[Thoughts]

这其实是一道数学题。题目已经说了,相同task之间要间隔n个interval,这就意味着任意一个循环子区间的长度是N+1.

这道题就简单了,统计在task表里出现次数最多的task是哪一个,出现了多少次。这里假设是A和C,分别出现了m次。那么我们肯定知道会有 m-1个循环子区间,而每个子区间的长度是N+1(task不够就必须用idle来补上)。而最后一个区间只有A和C各出现一次。


[Code]

1:    int leastInterval(vector<char>& tasks, int n) {  
2:      unordered_map<char, int> mp;  
3:        
4:      int count =0;  
5:      for(auto t : tasks) {  
6:        mp[t]++;  
7:        count = max(count, mp[t]);  
8:      }  
9:        
10:      int ans = (count-1)*(n+1);  
11:      // in case multiple tasks hold the same max count  
12:      for(auto e : mp) {  
13:        if(e.second == count) ans++;  
14:      }  
15:        
16:      return max((int)tasks.size(), ans);  
17:    }  


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