[LeetCode] 895. Maximum Frequency Stack

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.

最大频率栈。

设计一个类似堆栈的数据结构，将元素推入堆栈，并从堆栈中弹出出现频率最高的元素。

实现 FreqStack 类:

FreqStack() 构造一个空的堆栈。
void push(int val) 将一个整数 val 压入栈顶。
int pop() 删除并返回堆栈中出现频率最高的元素。
如果出现频率最高的元素不只一个，则移除并返回最接近栈顶的元素。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/maximum-frequency-stack
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这是一道设计题。题意说的很明白了，我直接给思路。这里我需要

• 变量max，记录全局最多的出现次数是多少
• Hashmap<Integer, Integer> freq - 记录每个元素的出现次数
• HashMap<Integer, Stack<Integer>> map - 记录相同出现次数都有哪些元素，存在stack里面

当我们做push操作的时候，正常更新freq哈希表中每个不同元素的出现次数；对于map哈希表，我们也将当前push的这个元素放入他对应的出现次数背后的那个stack中。

当我们做pop操作的时候，正常更新freq哈希表中每个不同元素的出现次数；对于map哈希表，注意因为题目要求pop()函数弹出出现次数最多的元素，所以我们肯定是从map.get(max)这个key背后的stack中弹出元素。注意如果做完pop操作之后这个stack为空了，需要将max--。

时间O(1)

空间O(n)

Java实现

class FreqStack {
    // 记录元素的出现次数
    HashMap<Integer, Integer> freq;
    // 记录相同出现次数都有哪些元素，用stack存
    HashMap<Integer, Stack<Integer>> map;
    int max;
    
    public FreqStack() {
        freq = new HashMap<>();
        map = new HashMap<>();
        max = 0;
    }
    
    public void push(int val) {
        int f = freq.getOrDefault(val, 0) + 1;
        freq.put(val, f);
        max = Math.max(max, f);
        if (!map.containsKey(f)) {
            map.put(f, new Stack<>());
        }
        map.get(f).push(val);
    }
    
    public int pop() {
        int x = map.get(max).pop();
        freq.put(x, max - 1);
        if (map.get(max).size() == 0) {
            max--;
        }
        return x;
    }
}

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

 

LeetCode 题目总结