[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:    }  

[Leetcode] Valid Anagram, Solution

Given two strings s and t, write a function to determine if t is an anagram of s.

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.

Note:
You may assume the string contains only lowercase alphabets.

[Thoughts]

对于字符出现次数做个统计就好了。因为只有26个小写字母，所以可以建立一个大小为26的索引数组charcount，用来统计每个字符的出现次数。

对于s, 将其作为字符数组进行遍历，在遍历的过程中，对每个出现的字符计数加一。

对于t， 同样将其遍历，对每个出现的字符计数减一。

如果s和t是anagram ， 那么最后的charcount数组中所有字符的计数都应该是0， 否则就不是anagram。

[Code]

1:  class Solution {  
2:  public:  
3:    bool isAnagram(string s, string t) {  
4:      vector<int> charcount(26, 0);  
5:      for(int i =0; i< s.length(); i++) {  
6:        charcount[s[i] - 'a'] ++;  
7:      }  
8:      for(int i =0; i< t.length(); i++) {  
9:        charcount[t[i] - 'a'] --;  
10:      }  
11:      for(int i =0; i<charcount.size(); i++) {  
12:        if(charcount[i] != 0) {  
13:          return false;  
14:        }  
15:      }  
16:      return true;  
17:    }  
18:  };  

Github: https://github.com/codingtmd/leetcode/blob/master/src/Valid%20Anagram.cpp