mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-11-04 03:33:12 +08:00
45 lines
1.4 KiB
HTML
45 lines
1.4 KiB
HTML
<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> </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(); --> 返回 -3.
|
||
minStack.pop();
|
||
minStack.top(); --> 返回 0.
|
||
minStack.getMin(); --> 返回 -2.
|
||
</pre>
|
||
|
||
<p> </p>
|
||
|
||
<p><strong>提示：</strong></p>
|
||
|
||
<ul>
|
||
<li><code>-2<sup>31</sup> <= val <= 2<sup>31</sup> - 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>, <code>pop</code>, <code>top</code>, and <code>getMin</code>最多被调用 <code>3 * 10<sup>4</sup></code> 次</li>
|
||
</ul>
|