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52 lines
2.0 KiB
HTML
52 lines
2.0 KiB
HTML
<p>请你实现一个「数字乘积类」<code>ProductOfNumbers</code>，要求支持下述两种方法：</p>
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<p>1.<code> add(int num)</code></p>
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<ul>
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<li>将数字 <code>num</code> 添加到当前数字列表的最后面。</li>
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</ul>
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<p>2.<code> getProduct(int k)</code></p>
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<ul>
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<li>返回当前数字列表中，最后 <code>k</code> 个数字的乘积。</li>
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<li>你可以假设当前列表中始终 <strong>至少</strong> 包含 <code>k</code> 个数字。</li>
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</ul>
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<p>题目数据保证：任何时候，任一连续数字序列的乘积都在 32-bit 整数范围内，不会溢出。</p>
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<p> </p>
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<p><strong>示例：</strong></p>
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<pre><strong>输入：</strong>
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["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"]
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[[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]
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<strong>输出：</strong>
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[null,null,null,null,null,null,20,40,0,null,32]
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<strong>解释：</strong>
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ProductOfNumbers productOfNumbers = new ProductOfNumbers();
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productOfNumbers.add(3); // [3]
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productOfNumbers.add(0); // [3,0]
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productOfNumbers.add(2); // [3,0,2]
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productOfNumbers.add(5); // [3,0,2,5]
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productOfNumbers.add(4); // [3,0,2,5,4]
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productOfNumbers.getProduct(2); // 返回 20 。最后 2 个数字的乘积是 5 * 4 = 20
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productOfNumbers.getProduct(3); // 返回 40 。最后 3 个数字的乘积是 2 * 5 * 4 = 40
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productOfNumbers.getProduct(4); // 返回 0 。最后 4 个数字的乘积是 0 * 2 * 5 * 4 = 0
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productOfNumbers.add(8); // [3,0,2,5,4,8]
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productOfNumbers.getProduct(2); // 返回 32 。最后 2 个数字的乘积是 4 * 8 = 32
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>add</code> 和 <code>getProduct</code> 两种操作加起来总共不会超过 <code>40000</code> 次。</li>
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<li><code>0 <= num <= 100</code></li>
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<li><code>1 <= k <= 40000</code></li>
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</ul>
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