maximum-frequency-stack

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Algorithms

Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.

Implement the <code>FreqStack</code> class:

<ul>
	<li><code>FreqStack()</code> constructs an empty frequency stack.</li>
	<li><code>void push(int val)</code> pushes an integer <code>val</code> onto the top of the stack.</li>
	<li><code>int pop()</code> removes and returns the most frequent element in the stack.
	<ul>
		<li>If there is a tie for the most frequent element, the element closest to the stack&#39;s top is removed and returned.</li>
	</ul>
	</li>
</ul>

&nbsp;
Example 1:

<pre>
Input
[&quot;FreqStack&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;]
[[], [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].
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>0 &lt;= val &lt;= 109</code></li>
	<li>At most <code>2 * 104</code> calls will be made to <code>push</code> and <code>pop</code>.</li>
	<li>It is guaranteed that there will be at least one element in the stack before calling <code>pop</code>.</li>
</ul>

Maximum Frequency Stack

Stack

Design

Hash Table

Ordered Set

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 <code>FreqStack</code>&nbsp;类:

<ul>
	<li><meta charset="UTF-8" /><code>FreqStack()</code>&nbsp;构造一个空的堆栈。</li>
	<li><meta charset="UTF-8" /><code>void push(int val)</code>&nbsp;将一个整数&nbsp;<code>val</code>&nbsp;压入栈顶。</li>
	<li><meta charset="UTF-8" /><code>int pop()</code>&nbsp;删除并返回堆栈中出现频率最高的元素。
	<ul>
		<li>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。</li>
	</ul>
	</li>
</ul>

&nbsp;

示例 1：

<pre>
输入：
["FreqStack","push","push","push","push","push","push","pop","pop","pop","pop"],
[[],[5],[7],[5],[7],[4],[5],[],[],[],[]]
输出：[null,null,null,null,null,null,null,5,7,5,4]
解释：
FreqStack = new FreqStack();
freqStack.push (5);//堆栈为 [5]
freqStack.push (7);//堆栈是 [5,7]
freqStack.push (5);//堆栈是 [5,7,5]
freqStack.push (7);//堆栈是 [5,7,5,7]
freqStack.push (4);//堆栈是 [5,7,5,7,4]
freqStack.push (5);//堆栈是 [5,7,5,7,4,5]
freqStack.pop ();//返回 5 ，因为 5 出现频率最高。堆栈变成 [5,7,5,7,4]。
freqStack.pop ();//返回 7 ，因为 5 和 7 出现频率最高，但7最接近顶部。堆栈变成 [5,7,5,4]。
freqStack.pop ();//返回 5 ，因为 5 出现频率最高。堆栈变成 [5,7,4]。
freqStack.pop ();//返回 4 ，因为 4, 5 和 7 出现频率最高，但 4 是最接近顶部的。堆栈变成 [5,7]。</pre>

&nbsp;

提示：

<ul>
	<li><code>0 &lt;= val &lt;= 109</code></li>
	<li><code>push</code>&nbsp;和 <code>pop</code>&nbsp;的操作数不大于 <code>2 * 104</code>。</li>
	<li>输入保证在调用&nbsp;<code>pop</code>&nbsp;之前堆栈中至少有一个元素。</li>
</ul>