Leetcode 之 Exclusive Time of Functions

636. Exclusive Time of Functions

1.Problem

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 = 2
logs = 
["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

2.Solution

logs数组中的一条数据代表：functionID + type + timesliceID ,stack中存functionID，因为函数是递归调用的，完全可以用栈来模拟。prev用来标记上一个函数开始执行时的时间片ID(timesliceID)
　　遍历logs(List)，对取出的每一个字符串进行分割成functionID ，type 和 timesliceID
       IF type == “start”
　　　　　　　IF 栈不空
　　　　　　　　　　timesliceID - prev 累加到栈顶元素的执行时间上去。
　　　　　　  将functionID 压栈；
　　　　　　　更新prev = timesliceID；
       ELSE type == “end”
　　　　　　 弹栈，取得一个functionID，更新这个functionID的执行时间为 （累加） timesliceID - prev + 1; (为什么 + 1 ，因为 end指令 那一个时间片函数依然再执行，也是函数执行的最后一个时间片)
　　　　　　 prev = timesliceID + 1； 同样的道理，这个timeliceID结束后，才开始执行别的函数

3.Code

 

class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        Stack<Integer> stack = new Stack<>();
        //Initialize
        int[] res = new int[n];
        int prevous = 0;
        for ( String s : logs ) {
            String[] ss = s.split(":");
            if ( ss[1].equals("start") ) {
                if ( !stack.empty() ) {
                    res[stack.peek()] +=  Integer.parseInt(ss[2]) - prevous;
                }
                    stack.push(Integer.parseInt(ss[0]));
                    prevous = Integer.parseInt(ss[2]);
            } else {
                res[Integer.parseInt(ss[0])] += Integer.parseInt(ss[2]) - prevous + 1;
                stack.pop();
                prevous = Integer.parseInt(ss[2]) + 1;
            }
        }
        return res;        
    }
}

 

4.提交Leetcode的代码

 

转载于:https://www.cnblogs.com/fangpengchengbupter/p/7422687.html