Skip to content

895. Maximum Frequency Stack

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.

Solution:

class FreqStack {
    Map<Integer, Integer> freqMap; // 记录每个值的频率
    Map<Integer, Stack<Integer>> groupMap; // 按频率分组存储值
    int maxFreq; // 当前的最大频率

    public FreqStack() {
        freqMap = new HashMap<>();
        groupMap = new HashMap<>();
        maxFreq = 0;
    }

    public void push(int val) {
        int freq = freqMap.getOrDefault(val, 0) + 1; // 增加频率
        freqMap.put(val, freq); // 更新频率
        maxFreq = Math.max(maxFreq, freq); // 更新最大频率
        groupMap.computeIfAbsent(freq, z -> new Stack<>()).push(val); // 将值加入对应频率的栈
    }

    public int pop() {
        int val = groupMap.get(maxFreq).pop(); // 从最大频率栈中弹出一个值
        freqMap.put(val, freqMap.get(val) - 1); // 减少频率
        if (groupMap.get(maxFreq).isEmpty()) { // 如果当前最大频率栈为空
            maxFreq--; // 更新最大频率
        }
        return val;
    }
}
//????