Description

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

Solution

Stack, time O(n), space O(n)

首先吐槽下題目的長度。這道題的input是保證valid的，那么可以用Stack記錄所有unfinished function calls，然后記錄每個(gè)時(shí)間片對應(yīng)的function和time即可。

注意必須一個(gè)時(shí)間片一個(gè)時(shí)間片計(jì)算，必須在三個(gè)start task a,b,c先后進(jìn)棧時(shí)，b進(jìn)棧時(shí)必須更新a占用的時(shí)間，否則等b出棧時(shí)計(jì)算a的時(shí)間就會有問題，因?yàn)橐呀?jīng)出棧的c的信息會被遺忘。

class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] times = new int[n];
        Stack<int[]> stack = new Stack<>();
        
        for (String log : logs) {
            String[] arr = log.split(":");
            int id = Integer.parseInt(arr[0]);
            boolean isStart = "start".equals(arr[1]);
            int time = Integer.parseInt(arr[2]);
            
            if (isStart) {
                if (!stack.empty()) {
                    times[stack.peek()[0]] += time - stack.peek()[1];  // important!
                }
                stack.push(new int[] {id, time});
            } else {
                int[] curr = stack.pop();
                int startTime = curr[1];
                times[id] += time + 1 - startTime;
                
                if (!stack.empty()) {
                    int[] pre = stack.pop();
                    pre[1] = time + 1;
                    stack.push(pre);
                }
            }
        }
        
        return times;
    }
}

簡化成下面這樣更好：

class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] times = new int[n];
        Stack<Integer> stack = new Stack<>();   // store unfinished function calls
        int preTime = 0;
        int runningFunc = 0;
        
        for (String log : logs) {
            String[] arr = log.split(":");
            int func = Integer.parseInt(arr[0]);
            boolean isStart = "start".equals(arr[1]);
            int time = Integer.parseInt(arr[2]);
            time += isStart ? 0 : 1;  // because end time is inclusive
            
            times[runningFunc] += time - preTime;
            
            if (isStart) {
                stack.push(runningFunc);
                runningFunc = func;
            } else {
                runningFunc = stack.pop();
            }
            
            preTime = time;
        }
        
        return times;
    }
}