Maximum Frequency Stack

Implement FreqStack, a class which simulates the operation of a stack-like data structure.

FreqStack has two functions:

• push(int x), which pushes an integer x onto the stack.
• pop(), which removes and returns the most frequent element in the stack.
  • If there is a tie for most frequent element, the element closest to the top of the stack is removed and returned. 

Example 1:

Input: 
["FreqStack","push","push","push","push","push","push","pop","pop","pop","pop"],
[[],[5],[7],[5],[7],[4],[5],[],[],[],[]]
Output: [null,null,null,null,null,null,null,5,7,5,4]
Explanation:
After making six .push operations, the stack is [5,7,5,7,4,5] from bottom to top.  Then:

pop() -> returns 5, as 5 is the most frequent.
The stack becomes [5,7,5,7,4].

pop() -> returns 7, as 5 and 7 is the most frequent, but 7 is closest to the top.
The stack becomes [5,7,5,4].

pop() -> returns 5.
The stack becomes [5,7,4].

pop() -> returns 4.
The stack becomes [5,7].

Note:

• Calls to FreqStack.push(int x) will be such that 0 <= x <= 10^9.
• It is guaranteed that FreqStack.pop() won't be called if the stack has zero elements.
• The total number of FreqStack.push calls will not exceed 10000 in a single test case.
• The total number of FreqStack.pop calls will not exceed 10000in a single test case.
• The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.

分析：

因为pop()一定是pop out频率最高，并且位置最靠近stack顶端的,所以，我们可以创建HashMap<Integer, Stack<Integer>> 这样一个map，key是频率，值是同一频率的值。

我们每次push的时候，需要知道那个值当前的频率，所以，我们需要有一个map来保存值与频率的关系。

 1 class FreqStack {
 2     HashMap<Integer, Integer> freq = new HashMap<>();
 3     HashMap<Integer, Stack<Integer>> m = new HashMap<>();
 4     int maxfreq = 0;
 5 
 6     public void push(int x) {
 7         int f = freq.getOrDefault(x, 0) + 1;
 8         freq.put(x, f);
 9         maxfreq = Math.max(maxfreq, f);
10         if (!m.containsKey(f)) m.put(f, new Stack<Integer>());
11         m.get(f).add(x);
12     }
13 
14     public int pop() {
15         int x = m.get(maxfreq).pop();
16         freq.put(x, maxfreq - 1);
17         if (m.get(maxfreq).size() == 0) maxfreq--;
18         return x;
19     }
20 }

 

转载于:https://www.cnblogs.com/beiyeqingteng/p/11213403.html