{ "data": { "question": { "questionId": "636", "questionFrontendId": "636", "categoryTitle": "Algorithms", "boundTopicId": 1805, "title": "Exclusive Time of Functions", "titleSlug": "exclusive-time-of-functions", "content": "

On a single-threaded CPU, we execute a program containing n functions. Each function has a unique ID between 0 and n-1.

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Function calls are stored in a call stack: 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 the current function being executed. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp.

\n\n

You are given a list logs, where logs[i] represents the ith log message formatted as a string "{function_id}:{"start" | "end"}:{timestamp}". For example, "0:start:3" means a function call with function ID 0 started at the beginning of timestamp 3, and "1:end:2" means a function call with function ID 1 ended at the end of timestamp 2. Note that a function can be called multiple times, possibly recursively.

\n\n

A function's exclusive time 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 2 time units and another call executing for 1 time unit, the exclusive time is 2 + 1 = 3.

\n\n

Return the exclusive time of each function in an array, where the value at the ith index represents the exclusive time for the function with ID i.

\n\n

 

\n

Example 1:

\n\"\"\n
\nInput: n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"]\nOutput: [3,4]\nExplanation:\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
\n\n

Example 2:

\n\n
\nInput: n = 1, logs = ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"]\nOutput: [8]\nExplanation:\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
\n\n

Example 3:

\n\n
\nInput: n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:6","1:end:6","0:end:7"]\nOutput: [7,1]\nExplanation:\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
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "函数的独占时间", "translatedContent": "

有一个 单线程 CPU 正在运行一个含有 n 道函数的程序。每道函数都有一个位于  0 和 n-1 之间的唯一标识符。

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函数调用 存储在一个 调用栈 上 ：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 当前正在执行的函数 。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。

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给你一个由日志组成的列表 logs ，其中 logs[i] 表示第 i 条日志消息，该消息是一个按 \"{function_id}:{\"start\" | \"end\"}:{timestamp}\" 进行格式化的字符串。例如，\"0:start:3\" 意味着标识符为 0 的函数调用在时间戳 3 的 起始开始执行 ；而 \"1:end:2\" 意味着标识符为 1 的函数调用在时间戳 2 的 末尾结束执行。注意，函数可以 调用多次，可能存在递归调用 。

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函数的 独占时间 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 2 单位时间，另一次调用执行 1 单位时间，那么该函数的 独占时间 为 2 + 1 = 3 。

\n\n

以数组形式返回每个函数的 独占时间 ，其中第 i 个下标对应的值表示标识符 i 的函数的独占时间。

\n \n\n

示例 1：

\n\"\"\n
\n输入：n = 2, logs = [\"0:start:0\",\"1:start:2\",\"1:end:5\",\"0:end:6\"]\n输出：[3,4]\n解释：\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
\n\n

示例 2：

\n\n
\n输入：n = 1, logs = [\"0:start:0\",\"0:start:2\",\"0:end:5\",\"0:start:6\",\"0:end:6\",\"0:end:7\"]\n输出：[8]\n解释：\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
\n\n

示例 3：

\n\n
\n输入：n = 2, logs = [\"0:start:0\",\"0:start:2\",\"0:end:5\",\"1:start:6\",\"1:end:6\",\"0:end:7\"]\n输出：[7,1]\n解释：\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。
\n\n

示例 4：

\n\n
\n输入：n = 2, logs = [\"0:start:0\",\"0:start:2\",\"0:end:5\",\"1:start:7\",\"1:end:7\",\"0:end:8\"]\n输出：[8,1]\n
\n\n

示例 5：

\n\n
\n输入：n = 1, logs = [\"0:start:0\",\"0:end:0\"]\n输出：[1]\n
\n\n

 

\n\n

提示：

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