895. Maximum Frequency Stack

题目

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.

题目大意

实现 FreqStack，模拟类似栈的数据结构的操作的一个类。

FreqStack 有两个函数：

• push(int x)，将整数 x 推入栈中。
• pop()，它移除并返回栈中出现最频繁的元素。如果最频繁的元素不只一个，则移除并返回最接近栈顶的元素。

解题思路

FreqStack 里面保存频次的 map 和相同频次 group 的 map。push 的时候动态的维护 x 的频次，并更新到对应频次的 group 中。pop 的时候对应减少频次字典里面的频次，并更新到对应频次的 group 中。

代码

package leetcode

type FreqStack struct {
	freq    map[int]int
	group   map[int][]int
	maxfreq int
}

func Constructor895() FreqStack {
	hash := make(map[int]int)
	maxHash := make(map[int][]int)
	return FreqStack{freq: hash, group: maxHash}
}

func (this *FreqStack) Push(x int) {
	if _, ok := this.freq[x]; ok {
		this.freq[x]++
	} else {
		this.freq[x] = 1
	}
	f := this.freq[x]
	if f > this.maxfreq {
		this.maxfreq = f
	}

	this.group[f] = append(this.group[f], x)
}

func (this *FreqStack) Pop() int {
	tmp := this.group[this.maxfreq]
	x := tmp[len(tmp)-1]
	this.group[this.maxfreq] = this.group[this.maxfreq][:len(this.group[this.maxfreq])-1]
	this.freq[x]--
	if len(this.group[this.maxfreq]) == 0 {
		this.maxfreq--
	}
	return x
}

/**
 * Your FreqStack object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Push(x);
 * param_2 := obj.Pop();
 */

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