51 lines
3.1 KiB
Markdown
51 lines
3.1 KiB
Markdown
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<p>创建一个基于时间的键值存储类 <code>TimeMap</code>,它支持下面两个操作:</p>
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<p>1. <code>set(string key, string value, int timestamp)</code></p>
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<ul>
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<li>存储键 <code>key</code>、值 <code>value</code>,以及给定的时间戳 <code>timestamp</code>。</li>
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</ul>
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<p>2. <code>get(string key, int timestamp)</code></p>
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<ul>
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<li>返回先前调用 <code>set(key, value, timestamp_prev)</code> 所存储的值,其中 <code>timestamp_prev <= timestamp</code>。</li>
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<li>如果有多个这样的值,则返回对应最大的 <code>timestamp_prev</code> 的那个值。</li>
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<li>如果没有值,则返回空字符串(<code>""</code>)。</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><strong>输入:</strong>inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
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<strong>输出:</strong>[null,null,"bar","bar",null,"bar2","bar2"]
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<strong>解释:</strong>
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TimeMap kv;
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kv.set("foo", "bar", 1); // 存储键 "foo" 和值 "bar" 以及时间戳 timestamp = 1
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kv.get("foo", 1); // 输出 "bar"
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kv.get("foo", 3); // 输出 "bar" 因为在时间戳 3 和时间戳 2 处没有对应 "foo" 的值,所以唯一的值位于时间戳 1 处(即 "bar")
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kv.set("foo", "bar2", 4);
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kv.get("foo", 4); // 输出 "bar2"
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kv.get("foo", 5); // 输出 "bar2"
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
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<strong>输出:</strong>[null,null,null,"","high","high","low","low"]
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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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<ol>
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<li>所有的键/值字符串都是小写的。</li>
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<li>所有的键/值字符串长度都在 <code>[1, 100]</code> 范围内。</li>
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<li>所有 <code>TimeMap.set</code> 操作中的时间戳 <code>timestamps</code> 都是严格递增的。</li>
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<li><code>1 <= timestamp <= 10^7</code></li>
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<li><code>TimeMap.set</code> 和 <code>TimeMap.get</code> 函数在每个测试用例中将(组合)调用总计 <code>120000</code> 次。</li>
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</ol>
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<div><div>Related Topics</div><div><li>设计</li><li>哈希表</li><li>字符串</li><li>二分查找</li></div></div>\n<div><li>👍 120</li><li>👎 0</li></div>
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