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存量题库数据更新

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小墨
2023-12-09 18:42:21 +08:00
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leetcode/problem/exclusive-time-of-functions.html
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@@ -9,7 +9,7 @@
<p>Return <em>the <strong>exclusive time</strong> of each function in an array, where the value at the </em><code>i<sup>th</sup></code><em> index represents the exclusive time for the function with ID </em><code>i</code>.</p>
<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/04/05/diag1b.png" style="width: 550px; height: 239px;" />
<pre>
<strong>Input:</strong> n = 2, logs = [&quot;0:start:0&quot;,&quot;1:start:2&quot;,&quot;1:end:5&quot;,&quot;0:end:6&quot;]
@@ -21,7 +21,7 @@ Function 0 resumes execution at the beginning of time 6 and executes for 1 unit
So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.
</pre>
<p><strong>Example 2:</strong></p>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> n = 1, logs = [&quot;0:start:0&quot;,&quot;0:start:2&quot;,&quot;0:end:5&quot;,&quot;0:start:6&quot;,&quot;0:end:6&quot;,&quot;0:end:7&quot;]
@@ -35,7 +35,7 @@ Function 0 (initial call) resumes execution at the beginning of time 7 and execu
So function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing.
</pre>
<p><strong>Example 3:</strong></p>
<p><strong class="example">Example 3:</strong></p>
<pre>
<strong>Input:</strong> n = 2, logs = [&quot;0:start:0&quot;,&quot;0:start:2&quot;,&quot;0:end:5&quot;,&quot;1:start:6&quot;,&quot;1:end:6&quot;,&quot;0:end:7&quot;]