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LeetCode 895. Maximum Frequency Stack

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题目

原题在此

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 10000 in a single test case.
  • The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.

解析

设计数据结构的题, 选对基本数据结构是关键. 要pop频率最高的元素, 就要统计每个元素出现频率. 用map<element, freq>是个思路; 其次在元素频率相同时, pop最近一次push的元素, 将相同freq的元素放入同一个stack即可 => map<freq, stack<element>>. 类成员变量里定义一个maxFreq, 在出退栈时更新.

代码

c++

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class FreqStack {
public:
    unordered_map<int, int> freq;
    unordered_map<int, stack<int>> m;
    int maxfreq = 0;
    FreqStack() {}

    // ++i will increment the value of i, and then return the incremented value.
    // i++ will increment the value of i, but return the original value that i held before being incremented.
    void push(int x) {
        maxfreq = max(maxfreq, ++freq[x]);
        m[freq[x]].push(x);
    }

    int pop() {
        int x = m[maxfreq].top();
        m[maxfreq].pop();
        if (!m[freq[x]--].size()) maxfreq--;
        return x;
    }
};