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https://gitee.com/coder-xiaomo/leetcode-problemset
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174 lines
29 KiB
JSON
174 lines
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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 class=\"example\">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 class=\"example\">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 class=\"example\">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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"translatedTitle": "函数的独占时间",
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"translatedContent": "<p>有一个 <strong>单线程</strong> CPU 正在运行一个含有 <code>n</code> 道函数的程序。每道函数都有一个位于 <code>0</code> 和 <code>n-1</code> 之间的唯一标识符。</p>\n\n<p>函数调用 <strong>存储在一个 <a href=\"https://baike.baidu.com/item/%E8%B0%83%E7%94%A8%E6%A0%88/22718047?fr=aladdin\" target=\"_blank\">调用栈</a> 上</strong> :当一个函数调用开始时,它的标识符将会推入栈中。而当一个函数调用结束时,它的标识符将会从栈中弹出。标识符位于栈顶的函数是 <strong>当前正在执行的函数</strong> 。每当一个函数开始或者结束时,将会记录一条日志,包括函数标识符、是开始还是结束、以及相应的时间戳。</p>\n\n<p>给你一个由日志组成的列表 <code>logs</code> ,其中 <code>logs[i]</code> 表示第 <code>i</code> 条日志消息,该消息是一个按 <code>\"{function_id}:{\"start\" | \"end\"}:{timestamp}\"</code> 进行格式化的字符串。例如,<code>\"0:start:3\"</code> 意味着标识符为 <code>0</code> 的函数调用在时间戳 <code>3</code> 的 <strong>起始开始执行</strong> ;而 <code>\"1:end:2\"</code> 意味着标识符为 <code>1</code> 的函数调用在时间戳 <code>2</code> 的 <strong>末尾结束执行</strong>。注意,函数可以 <strong>调用多次,可能存在递归调用 </strong>。</p>\n\n<p>函数的 <strong>独占时间</strong> 定义是在这个函数在程序所有函数调用中执行时间的总和,调用其他函数花费的时间不算该函数的独占时间。例如,如果一个函数被调用两次,一次调用执行 <code>2</code> 单位时间,另一次调用执行 <code>1</code> 单位时间,那么该函数的 <strong>独占时间</strong> 为 <code>2 + 1 = 3</code> 。</p>\n\n<p>以数组形式返回每个函数的 <strong>独占时间</strong> ,其中第 <code>i</code> 个下标对应的值表示标识符 <code>i</code> 的函数的独占时间。</p>\n \n\n<p><strong>示例 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>输入:</strong>n = 2, logs = [\"0:start:0\",\"1:start:2\",\"1:end:5\",\"0:end:6\"]\n<strong>输出:</strong>[3,4]\n<strong>解释:</strong>\n函数 0 在时间戳 0 的起始开始执行,执行 2 个单位时间,于时间戳 1 的末尾结束执行。 \n函数 1 在时间戳 2 的起始开始执行,执行 4 个单位时间,于时间戳 5 的末尾结束执行。 \n函数 0 在时间戳 6 的开始恢复执行,执行 1 个单位时间。 \n所以函数 0 总共执行 2 + 1 = 3 个单位时间,函数 1 总共执行 4 个单位时间。 \n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 1, logs = [\"0:start:0\",\"0:start:2\",\"0:end:5\",\"0:start:6\",\"0:end:6\",\"0:end:7\"]\n<strong>输出:</strong>[8]\n<strong>解释:</strong>\n函数 0 在时间戳 0 的起始开始执行,执行 2 个单位时间,并递归调用它自身。\n函数 0(递归调用)在时间戳 2 的起始开始执行,执行 4 个单位时间。\n函数 0(初始调用)恢复执行,并立刻再次调用它自身。\n函数 0(第二次递归调用)在时间戳 6 的起始开始执行,执行 1 个单位时间。\n函数 0(初始调用)在时间戳 7 的起始恢复执行,执行 1 个单位时间。\n所以函数 0 总共执行 2 + 4 + 1 + 1 = 8 个单位时间。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 2, logs = [\"0:start:0\",\"0:start:2\",\"0:end:5\",\"1:start:6\",\"1:end:6\",\"0:end:7\"]\n<strong>输出:</strong>[7,1]\n<strong>解释:</strong>\n函数 0 在时间戳 0 的起始开始执行,执行 2 个单位时间,并递归调用它自身。\n函数 0(递归调用)在时间戳 2 的起始开始执行,执行 4 个单位时间。\n函数 0(初始调用)恢复执行,并立刻调用函数 1 。\n函数 1在时间戳 6 的起始开始执行,执行 1 个单位时间,于时间戳 6 的末尾结束执行。\n函数 0(初始调用)在时间戳 7 的起始恢复执行,执行 1 个单位时间,于时间戳 7 的末尾结束执行。\n所以函数 0 总共执行 2 + 4 + 1 = 7 个单位时间,函数 1 总共执行 1 个单位时间。 </pre>\n\n<p><strong>示例 4:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 2, logs = [\"0:start:0\",\"0:start:2\",\"0:end:5\",\"1:start:7\",\"1:end:7\",\"0:end:8\"]\n<strong>输出:</strong>[8,1]\n</pre>\n\n<p><strong>示例 5:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 1, logs = [\"0:start:0\",\"0:end:0\"]\n<strong>输出:</strong>[1]\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</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>两个开始事件不会在同一时间戳发生</li>\n\t<li>两个结束事件不会在同一时间戳发生</li>\n\t<li>每道函数都有一个对应 <code>\"start\"</code> 日志的 <code>\"end\"</code> 日志</li>\n</ul>\n",
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