Exclusive Time of Functions

https://leetcode.com/problems/exclusive-time-of-functions/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:

Input logs will be sorted by timestamp, NOT log id.

Your output should be sorted by function id, which means the 0th element of your output corresponds to the exclusive time of function 0.

Two functions won't start or end at the same time.

Functions could be called recursively, and will always end.

1 <= n <= 100

Thoughts

给一系列start和end ts表示函数开始和结束时间，让模拟调用栈 (call stack), 计算每个函数自身的执行时间。每个ts看成一个格子，start是格子起始，end是格子终止，也就是下个格子的起始。每个ts统一转化成起始时间，用pre_t记录上个ts，用t - pre_t即持续时长。

Code

/*
 * @lc app=leetcode id=636 lang=cpp
 *
 * [636] Exclusive Time of Functions
 */

// @lc code=start
class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        const auto split = [](const string &s, char delim) {
            vector<string> result;
            stringstream ss (s);
            string item;
            while (getline (ss, item, delim)) {
                result.push_back (item);
            }
            return result;
        };
        vector<int> res(n, 0);
        stack<int> st;
        int pre_t = 0;
        st.push(0);
        for (const auto &s : logs) {
            const auto &v = split(s, ':');
            const int id = stoi(v[0]);
            int t = stoi(v[2]);
            if (v[1] == "start") {
                res[st.top()] += t - pre_t;
                pre_t = t;
                st.push(id);
            } else {
                ++t;
                res[id] += t - pre_t;
                st.pop();
            }
            pre_t = t;
        }
        return res;
    }
};
// @lc code=end

class Solution {
    public int[] exclusiveTime(int n, List<String> logs) {
        int[] res = new int[n];
        Stack<Integer> stack = new Stack<>();
        int pre = 0; // pre means the start of the interval
        for (String log : logs) {
            String[] subs = log.split(":");
            int id = Integer.parseInt(subs[0]), time = Integer.parseInt(subs[2]);
            if(subs[1].equals("start")) {
                if(!stack.isEmpty())  res[stack.peek()] += time - pre;
                // time is the start of next interval, doesn't belong to current interval.
                stack.push(id);
                pre = time;
            } else {
                res[stack.pop()] += time - pre + 1;
                // time is end of current interval, belong to current interval. That's why we have +1 here
                pre = time + 1;
                // pre means the start of next interval, so we need to +1
            }
        }
        return res;
    }
}

Analysis

时间复杂度O(N).