LeetCode.636 Exclusive Time of Functions

题目：

Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find the exclusive time of these functions.

Each function has a unique id, start from 0 to n-1. A function may be called recursively or by another function.

A log is a string has this format : function_id:start_or_end:timestamp. For example, "0:start:0" means function 0 starts from the very beginning of time 0. "0:end:0" means function 0 ends to the very end of time 0.

Exclusive time of a function is defined as the time spent within this function, the time spent by calling other functions should not be considered as this function's exclusive time. You should return the exclusive time of each function sorted by their function id.

Example 1:

Input:n = 2logs = ["0:start:0", "1:start:2", "1:end:5", "0:end:6"]Output:[3, 4]Explanation:Function 0 starts at time 0, then it executes 2 units of time and reaches the end of time 1. Now function 0 calls function 1, function 1 starts at time 2, executes 4 units of time and end at time 5.Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. So function 0 totally execute 2 + 1 = 3 units of time, and function 1 totally execute 4 units of time.

Note:

1. Input logs will be sorted by timestamp, NOT log id.
2. Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.
3. Two functions won't start or end at the same time.
4. Functions could be called recursively, and will always end.
5. 1 <= n <= 100

分析：

class Solution {    public int[] exclusiveTime(int n, List<String> logs) {        //给定任务调度时间安排，返回各个进程运行的时间。        //注意：输入的调度根据时间排序，不是根据进程id。输出按照进程id输出。两个进程不会存在时间交叉。        //思路：使用一个数组来记录各进程的调度时间,两个进程实现记录各进程开始和结束时间，        //最后使用一个stack来记录运行先后顺序。类似甘特图。                        int[] res=new int[n];                Stack<Integer> stack=new Stack<Integer>();        //起始时间        int prevTime=0;        for(int i=0;i<logs.size();i++){            String str=logs.get(i);            String [] subStr=str.split(":");                        if(!stack.isEmpty()){                //取出元素                res[stack.peek()]+=Integer.parseInt(subStr[2])-prevTime;            }                        //更新起始时间            prevTime=Integer.parseInt(subStr[2]);                        if(subStr[1].equals("start")){                //开始                stack.push(Integer.parseInt(subStr[0]));            }else {                //结束.完整的进程不被打断，+1 例如2开始5结束 5-2+1=4；                res[stack.pop()]++;                //更新起始时间                prevTime++;            }        }        return res;    }}