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https://gitee.com/coder-xiaomo/leetcode-problemset
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165 lines
19 KiB
JSON
165 lines
19 KiB
JSON
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"title": "Exclusive Time of Functions",
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"content": "<p>On a <strong>single-threaded</strong> CPU, we execute a program containing <code>n</code> functions. Each function has a unique ID between <code>0</code> and <code>n-1</code>.</p>\n\n<p>Function calls are <strong>stored in a <a href=\"https://en.wikipedia.org/wiki/Call_stack\">call stack</a></strong>: when a function call starts, its ID is pushed onto the stack, and when a function call ends, its ID is popped off the stack. The function whose ID is at the top of the stack is <strong>the current function being executed</strong>. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp.</p>\n\n<p>You are given a list <code>logs</code>, where <code>logs[i]</code> represents the <code>i<sup>th</sup></code> log message formatted as a string <code>"{function_id}:{"start" | "end"}:{timestamp}"</code>. For example, <code>"0:start:3"</code> means a function call with function ID <code>0</code> <strong>started at the beginning</strong> of timestamp <code>3</code>, and <code>"1:end:2"</code> means a function call with function ID <code>1</code> <strong>ended at the end</strong> of timestamp <code>2</code>. Note that a function can be called <b>multiple times, possibly recursively</b>.</p>\n\n<p>A function's <strong>exclusive time</strong> is the sum of execution times for all function calls in the program. For example, if a function is called twice, one call executing for <code>2</code> time units and another call executing for <code>1</code> time unit, the <strong>exclusive time</strong> is <code>2 + 1 = 3</code>.</p>\n\n<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>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/04/05/diag1b.png\" style=\"width: 550px; height: 239px;\" />\n<pre>\n<strong>Input:</strong> n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"]\n<strong>Output:</strong> [3,4]\n<strong>Explanation:</strong>\nFunction 0 starts at the beginning of time 0, then it executes 2 for units of time and reaches the end of time 1.\nFunction 1 starts at the beginning of time 2, executes for 4 units of time, and ends at the end of time 5.\nFunction 0 resumes execution at the beginning of time 6 and executes for 1 unit of time.\nSo function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 1, logs = ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"]\n<strong>Output:</strong> [8]\n<strong>Explanation:</strong>\nFunction 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.\nFunction 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.\nFunction 0 (initial call) resumes execution then immediately calls itself again.\nFunction 0 (2nd recursive call) starts at the beginning of time 6 and executes for 1 unit of time.\nFunction 0 (initial call) resumes execution at the beginning of time 7 and executes for 1 unit of time.\nSo function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:6","1:end:6","0:end:7"]\n<strong>Output:</strong> [7,1]\n<strong>Explanation:</strong>\nFunction 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself.\nFunction 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time.\nFunction 0 (initial call) resumes execution then immediately calls function 1.\nFunction 1 starts at the beginning of time 6, executes 1 unit of time, and ends at the end of time 6.\nFunction 0 resumes execution at the beginning of time 6 and executes for 2 units of time.\nSo function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1 spends 1 unit of total time executing.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>1 <= logs.length <= 500</code></li>\n\t<li><code>0 <= function_id < n</code></li>\n\t<li><code>0 <= timestamp <= 10<sup>9</sup></code></li>\n\t<li>No two start events will happen at the same timestamp.</li>\n\t<li>No two end events will happen at the same timestamp.</li>\n\t<li>Each function has an <code>"end"</code> log for each <code>"start"</code> log.</li>\n</ul>\n",
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