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LeetCode: 895-Maximum Frequency Stack 解題紀錄

Last Updated on 2022-03-19 by Clay

題目

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.

題目要我們設計一個類似 stack 的結構,不過若要彈出pop),則必須要按照數值的頻率來排優先度;若是頻率一樣的數值,則像 stack 一樣越接近頂部越優先彈出。


解題思路

Hash Map 解法

老實說這題還滿容易就 TLE 了,所以比較簡單的做法應該是使用 hash map 來紀錄一個數值的出現次數、以及特定頻率下數值推入的順序。

如果彈出了數值,則需要把該數值對應的出現次數減去一、並將對應最高頻率的順序表彈出。

並且還需要一個紀錄最大出現次數的變數,並在彈出後檢查最大頻率的順序表是否為空;最大頻率對應的順序表若為空,則將最大頻率的變數減去一。

詳細的製作方法請參考程式碼。


C++ 範例程式碼

class FreqStack {
public:
    // Init
    int maxFreq = 0;
    unordered_map<int, int> val2freq;
    unordered_map<int, stack<int>> freq2stack;
    
    FreqStack() {
        
    }
    
    // Push09sxaz
    void push(int val) {
        ++val2freq[val];
        maxFreq = max(maxFreq, val2freq[val]);
        
        freq2stack[val2freq[val]].push(val);
    }
    
    int pop() {
        // Get the top value
        int top = freq2stack[maxFreq].top();
        freq2stack[maxFreq].pop();
        
        // Decrement the frquency of the top value (for the next push)
        --val2freq[top];
        
        // If the maxFreq stack has no any value
        if (freq2stack[maxFreq].empty()) --maxFreq;
        
        return top;
    }
};

/**
 * Your FreqStack object will be instantiated and called as such:
 * FreqStack* obj = new FreqStack();
 * obj->push(val);
 * int param_2 = obj->pop();
 */



兩節點 Python 範例程式碼

class FreqStack:
    def __init__(self):
        self.max_freq = 0
        self.val2freq = dict()
        self.freq2stack = dict()

    def push(self, val: int) -> None:
        freq = self.val2freq.get(val, 0) + 1
        
        self.val2freq[val] = freq
        self.max_freq = max(self.max_freq, freq)
        self.freq2stack[freq] = self.freq2stack.get(freq, []) + [val]

    def pop(self) -> int:
        top = self.freq2stack[self.max_freq].pop()
        self.val2freq[top] -= 1
        
        if len(self.freq2stack[self.max_freq]) == 0:
            self.max_freq -= 1
        
        return top
        


# Your FreqStack object will be instantiated and called as such:
# obj = FreqStack()
# obj.push(val)
# param_2 = obj.pop()

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