[LeetCode] 636. Exclusive Time of Functions
作者：互联网
On a single-threaded CPU, we execute a program containing n functions. Each function has a unique ID between 0 and n-1.
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.
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.
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.
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.
Example 1:
Input: n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"] Output: [3,4] Explanation: Function 0 starts at the beginning of time 0, then it executes 2 for units of time and reaches the end of time 1. Function 1 starts at the beginning of time 2, executes for 4 units of time, and ends at the end of time 5. Function 0 resumes execution at the beginning of time 6 and executes for 1 unit of time. So function 0 spends 2 + 1 = 3 units of total time executing, and function 1 spends 4 units of total time executing.
Example 2:
Input: n = 1, logs = ["0:start:0","0:start:2","0:end:5","0:start:6","0:end:6","0:end:7"] Output: [8] Explanation: Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself. Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time. Function 0 (initial call) resumes execution then immediately calls itself again. Function 0 (2nd recursive call) starts at the beginning of time 6 and executes for 1 unit of time. Function 0 (initial call) resumes execution at the beginning of time 7 and executes for 1 unit of time. So function 0 spends 2 + 4 + 1 + 1 = 8 units of total time executing.
Example 3:
Input: n = 2, logs = ["0:start:0","0:start:2","0:end:5","1:start:6","1:end:6","0:end:7"] Output: [7,1] Explanation: Function 0 starts at the beginning of time 0, executes for 2 units of time, and recursively calls itself. Function 0 (recursive call) starts at the beginning of time 2 and executes for 4 units of time. Function 0 (initial call) resumes execution then immediately calls function 1. Function 1 starts at the beginning of time 6, executes 1 unit of time, and ends at the end of time 6. Function 0 resumes execution at the beginning of time 6 and executes for 2 units of time. So function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1 spends 1 unit of total time executing.
Constraints:
1 <= n <= 1001 <= logs.length <= 5000 <= function_id < n0 <= timestamp <= 109- No two start events will happen at the same timestamp.
- No two end events will happen at the same timestamp.
- Each function has an
"end"log for each"start"log.
函数的独占时间。
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。
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。
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 。
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 。
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。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/exclusive-time-of-functions
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题意不难理解，思路是stack，理由是无论这里面涉及多少函数的递归调用，一定是后开始跑的函数先结束，符合stack的后进先出的原则。我们用题目给的第一个例子跑一下。这里我们还需要一个变量 last 记录上一个函数进出 stack 的时间。
- id = 每个函数的id
- time = 每个log里表示的时间戳
- operator = start/end
对于任何函数是否入栈，触发的条件是看 log 里是否出现 start 或者 end，无论出现哪一个，都是需要结算的时候。如果遇到某个 log 是 start，说明是某个函数的开始时间，此时如果栈不为空，说明栈顶那个函数需要暂停一下了，我们需要把此时栈顶函数已经花掉的时间先入栈（time - last），然后再把当前函数的id入栈。
如果遇到某个 log 是 end，就说明当前函数需要结算了，把栈顶元素弹出，此时这个函数的运行时间 = time - last + 1，并且要把 last 更新成 time + 1。
时间O(n)
空间O(n)
Java实现
1 class Solution {
2 public int[] exclusiveTime(int n, List<String> logs) {
3 int[] res = new int[n];
4 Deque<Integer> stack = new ArrayDeque<>();
5 int last = 0;
6 for (String log : logs) {
7 String[] strs = log.split("\\:");
8 int id = Integer.parseInt(strs[0]);
9 int time = Integer.parseInt(strs[2]);
10 String operator = strs[1];
11 if (operator.equals("start")) {
12 if (!stack.isEmpty()) {
13 res[stack.peek()] += time - last;
14 }
15 stack.push(id);
16 last = time;
17 } else {
18 res[stack.pop()] += time - last + 1;
19 last = time + 1;
20 }
21 }
22 return res;
23 }
24 }
标签：function,Exclusive,Functions,Function,end,函数,start,time,LeetCode 来源： https://www.cnblogs.com/cnoodle/p/16558451.html