LeetCode 895. Maximum Frequency Stack

原题链接在这里：https://leetcode.com/problems/maximum-frequency-stack/

题目：

Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.

Implement the FreqStack class:

• FreqStack() constructs an empty frequency stack.
• void push(int val) pushes an integer val onto the top of the stack.
• int pop() removes and returns the most frequent element in the stack.
  • If there is a tie for the most frequent element, the element closest to the stack’s top 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
FreqStack freqStack = new FreqStack();
freqStack.push(5); // The stack is [5]
freqStack.push(7); // The stack is [5,7]
freqStack.push(5); // The stack is [5,7,5]
freqStack.push(7); // The stack is [5,7,5,7]
freqStack.push(4); // The stack is [5,7,5,7,4]
freqStack.push(5); // The stack is [5,7,5,7,4,5]
freqStack.pop();   // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4].
freqStack.pop();   // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4].
freqStack.pop();   // return 5, as 5 is the most frequent. The stack becomes [5,7,4].
freqStack.pop();   // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].

Constraints:

• 0 <= val <= 109
• At most 2 * 104 calls will be made to push and pop.
• It is guaranteed that there will be at least one element in the stack before calling pop.

题解：

In order to popped the most frequent element and if ther are tie case pop the recent one.

We could have a Map<Integer, Stack<Integer>> group which records the frequency to the stack.

If x are pushed 3 times, then in group(1,2,3), each of them, the stack contains x.

Time Complexity: push, O(1). pop, O(1).

Space: O(n). n is the number of elements in group.

AC Java:

 1 class FreqStack {
 2     Map<Integer, Integer> freq;
 3     Map<Integer, Stack<Integer>> group;
 4     int max;
 5     
 6     public FreqStack() {
 7         freq = new HashMap<>();
 8         group = new HashMap<>();
 9         max = 0;    
10     }
11     
12     public void push(int val) {
13         int f = freq.getOrDefault(val, 0) + 1;
14         freq.put(val, f);
15         group.computeIfAbsent(f, z -> new Stack<Integer>()).push(val);
16         max = Math.max(max, f);
17     }
18     
19     public int pop() {
20         int res = group.get(max).pop();
21         freq.put(res, max - 1);
22         if(group.get(max).isEmpty()){
23             group.remove(max);
24             max--;
25         }
26         
27         return res;
28     }
29 }
30 
31 /**
32  * Your FreqStack object will be instantiated and called as such:
33  * FreqStack obj = new FreqStack();
34  * obj.push(val);
35  * int param_2 = obj.pop();
36  */