Project Notes

Min Stack

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Description

tab: English
 
<p>Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.</p>
 
<p>Implement the <code>MinStack</code> class:</p>
 
<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>
 
<p>You must implement a solution with <code>O(1)</code> time complexity for each function.</p>
 
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
 
<pre>
<strong>Input</strong>
[&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],[],[],[],[]]
 
<strong>Output</strong>
[null,null,null,null,-3,null,0,-2]
 
<strong>Explanation</strong>
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>
 
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
 
<ul>
	<li><code>-2<sup>31</sup> &lt;= val &lt;= 2<sup>31</sup> - 1</code></li>
	<li>Methods <code>pop</code>, <code>top</code> and <code>getMin</code> operations will always be called on <strong>non-empty</strong> stacks.</li>
	<li>At most <code>3 * 10<sup>4</sup></code> calls will be made to <code>push</code>, <code>pop</code>, <code>top</code>, and <code>getMin</code>.</li>
</ul>
 
 
 
> [!tip]- Hint 1
> 
> Consider each node in the stack having a minimum value. (Credits to @aakarshmadhavan)
 
 
---
 
[submissions](https://leetcode.com/problems/min-stack/submissions/) | [solutions](https://leetcode.com/problems/min-stack/solutions/)
 
 
tab: 中文
 
<p>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。</p>
 
<p>实现 <code>MinStack</code> 类:</p>
 
<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>
 
<p>&nbsp;</p>
 
<p><strong>示例 1:</strong></p>
 
<pre>
<strong>输入：</strong>
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
 
<strong>输出：</strong>
[null,null,null,null,-3,null,0,-2]
 
<strong>解释：</strong>
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>
 
<p>&nbsp;</p>
 
<p><strong>提示：</strong></p>
 
<ul>
	<li><code>-2<sup>31</sup>&nbsp;&lt;= val &lt;= 2<sup>31</sup>&nbsp;- 1</code></li>
	<li><code>pop</code>、<code>top</code> 和 <code>getMin</code> 操作总是在 <strong>非空栈</strong> 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 * 10<sup>4</sup></code>&nbsp;次</li>
</ul>
 
 
 
> [!tip]- 提示 1
> 
> Consider each node in the stack having a minimum value. (Credits to @aakarshmadhavan)
 
 
---
 
[提交记录](https://leetcode.cn/problems/min-stack/submissions/) | [题解](https://leetcode.cn/problems/min-stack/solution/)

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