Leetcode 895. Maximum Frequency Stack (Heap 题)

1. Maximum Frequency Stack
   Hard

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.

解法1：
用priority queue，默认就是最大堆。
这题要注意，当某个元素值(比如说5)多次出现时，每次push都是push一个新的节点，也就是push 5 3次，Node(5, 1, timestamp1), Node(5, 2, timestamp2), Node(5, 3, timestamp3)都会出现在heap中。这样，pop掉Node(5, 3, timestamp3)之后，Node(5, 2, timestamp2)还是存在heap中跟其他节点竞争。

struct Node {
    int val;
    int freq;
    int ts;
    Node(int v, int f, int t) : val(v), freq(f), ts(t) {}
    bool operator < (const Node & node) const {
        if (freq != node.freq) return freq < node.freq;
        return ts < node.ts;
    }
};
class FreqStack {
public:
    FreqStack() {
    }
    
    void push(int val) {
        timestamp++;
        val2freq[val]++;
        maxHeap.push(Node(val, val2freq[val], timestamp));
    }
    
    int pop() {
        timestamp++;
        Node top = maxHeap.top();
        maxHeap.pop();
        val2freq[top.val]--;
        return top.val;
    }

private:
    priority_queue<Node> maxHeap;
    map<int, int> val2freq;
    int timestamp = 0;
};

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