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54 lines
1.7 KiB
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54 lines
1.7 KiB
HTML
<p>请你设计一个 <strong>最小栈</strong> 。它提供 <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>
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<p> </p>
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<p>实现 <code>MinStack</code> 类:</p>
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<ul>
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<li><code>MinStack()</code> 初始化堆栈对象。</li>
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<li><code>void push(int val)</code> 将元素val推入堆栈。</li>
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<li><code>void pop()</code> 删除堆栈顶部的元素。</li>
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<li><code>int top()</code> 获取堆栈顶部的元素。</li>
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<li><code>int getMin()</code> 获取堆栈中的最小元素。</li>
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</ul>
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<p> </p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入：</strong>
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["MinStack","push","push","push","getMin","pop","top","getMin"]
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[[],[-2],[2],[-3],[],[],[],[]]
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<strong>输出：</strong>
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[null,null,null,null,-3,null,2,-2]
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<strong>解释：</strong>
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MinStack minStack = new MinStack();
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minStack.push(-2);
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minStack.push(2);
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minStack.push(-3);
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minStack.getMin(); --> 返回 -3.
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minStack.pop();
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minStack.top(); --> 返回 2.
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minStack.getMin(); --> 返回 -2.
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</pre>
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<p> </p>
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<p><strong> <br />
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提示：</strong></p>
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<ul>
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<li><code>-2<sup>31</sup> <= val <= 2<sup>31</sup> - 1</code></li>
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<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>
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<li><code>push</code>、<code>pop</code>、<code>top</code> 和 <code>getMin</code> 最多被调用 <code>3 * 10<sup>4</sup></code> 次</li>
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</ul>
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<p> </p>
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<p>注意：本题与主站 155 题相同：<a href="https://leetcode-cn.com/problems/min-stack/">https://leetcode-cn.com/problems/min-stack/</a></p>
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<p> </p>
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