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Algorithms

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

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

<ul>
	<li><code>MinStack()</code> initializes the stack object.</li>
	<li><code>void push(int val)</code> pushes the element <code>val</code> onto the stack.</li>
	<li><code>void pop()</code> removes the element on the top of the stack.</li>
	<li><code>int top()</code> gets the top element of the stack.</li>
	<li><code>int getMin()</code> retrieves the minimum element in the stack.</li>
</ul>

You must implement a solution with <code>O(1)</code> time complexity for each function.

&nbsp;
Example 1:

<pre>
Input
[&quot;MinStack&quot;,&quot;push&quot;,&quot;push&quot;,&quot;push&quot;,&quot;getMin&quot;,&quot;pop&quot;,&quot;top&quot;,&quot;getMin&quot;]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>-231 &lt;= val &lt;= 231 - 1</code></li>
	<li>Methods <code>pop</code>, <code>top</code> and <code>getMin</code> operations will always be called on non-empty stacks.</li>
	<li>At most <code>3 * 104</code> calls will be made to <code>push</code>, <code>pop</code>, <code>top</code>, and <code>getMin</code>.</li>
</ul>

Min Stack

Sliding Window Maximum

sliding-window-maximum

滑动窗口最大值

Max Stack

max-stack

最大栈

Stack

Design

设计一个支持 <code>push</code> ，<code>pop</code> ，<code>top</code> 操作，并能在常数时间内检索到最小元素的栈。

实现 <code>MinStack</code> 类:

<ul>
	<li><code>MinStack()</code> 初始化堆栈对象。</li>
	<li><code>void push(int val)</code> 将元素val推入堆栈。</li>
	<li><code>void pop()</code> 删除堆栈顶部的元素。</li>
	<li><code>int top()</code> 获取堆栈顶部的元素。</li>
	<li><code>int getMin()</code> 获取堆栈中的最小元素。</li>
</ul>

&nbsp;

示例 1:

<pre>
输入：
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

输出：
[null,null,null,null,-3,null,0,-2]

解释：
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --&gt; 返回 -3.
minStack.pop();
minStack.top(); --&gt; 返回 0.
minStack.getMin(); --&gt; 返回 -2.
</pre>

&nbsp;

提示：

<ul>
	<li><code>-231&nbsp;&lt;= val &lt;= 231&nbsp;- 1</code></li>
	<li><code>pop</code>、<code>top</code> 和 <code>getMin</code> 操作总是在 非空栈 上调用</li>
	<li><code>push</code>,&nbsp;<code>pop</code>,&nbsp;<code>top</code>, and&nbsp;<code>getMin</code>最多被调用&nbsp;<code>3 * 104</code>&nbsp;次</li>
</ul>