636. Exclusive Time of Functions (Medium)

Given the running logs ofnfunctions 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

其实就是简单的模拟call stack, 新函数的start()是老函数调用的, start并不会关闭上一个函数, 而是把新函数的局部变量压入call stack.

Solution 1: Stack O(n); O(n)

public int[] exclusiveTime(int n, List<String> logs) {
    // separate time to several intervals, add interval to their function
    int[] result = new int[n];
    Stack<Integer> st = new Stack<>();
    int pre = 0;
    // pre means the start of the interval
    for(String log: logs) {
        String[] arr = log.split(":");
        if(arr[1].equals("start")) {
            if(!st.isEmpty())  result[st.peek()] += Integer.parseInt(arr[2]) - pre;
            // arr[2] is the start of next interval, doesn't belong to current interval.
            st.push(Integer.parseInt(arr[0]));//碰到start就压栈
            pre = Integer.parseInt(arr[2]);//更新pre
        } else {
            result[st.pop()] += Integer.parseInt(arr[2]) - pre + 1;//碰到end就出栈
            // arr[2] is end of current interval, belong to current interval. That's why we have +1 here
            pre = Integer.parseInt(arr[2]) + 1;//更新pre，记得加1
            // pre means the start of next interval, so we need to +1
        }
    }
    return result;
}