[LeetCode] 636. Exclusive Time of Functions 函数的独家时间

2021年09月15日 阅读数:1
这篇文章主要向大家介绍[LeetCode] 636. Exclusive Time of Functions 函数的独家时间,主要内容包括基础应用、实用技巧、原理机制等方面,希望对大家有所帮助。

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

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

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.python

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.ios

Example 1:数组

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:app

  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

给一个n个函数运行记录的log数组,表示非抢占式CPU调用函数的状况,即同时只能运行一个函数。格式为id:start or end:time,求每一个函数占用cpu的总时间。ide

解法:栈,用stack保存当前还在还没结束的function的(id,start)。遇到是start的log时,计算栈顶函数的时间,而后把当前函数push进stack;遇到是end的就pop出stack的最近一个元素,并计算此function的运行时间。函数

Java:post

public int[] exclusiveTime(int n, List<String> logs) {
    int[] res = new int[n];
    Stack<Integer> stack = new Stack<>();
    int prevTime = 0;
    for (String log : logs) {
        String[] parts = log.split(":");
        if (!stack.isEmpty()) res[stack.peek()] +=  Integer.parseInt(parts[2]) - prevTime; 
        prevTime = Integer.parseInt(parts[2]);
        if (parts[1].equals("start")) stack.push(Integer.parseInt(parts[0]));
        else {
            res[stack.pop()]++;
            prevTime++;
        }
    }
    return res;
}  

Python:this

# Time:  O(n)
# Space: O(n)
class Solution(object):
    def exclusiveTime(self, n, logs):
        """
        :type n: int
        :type logs: List[str]
        :rtype: List[int]
        """
        result = [0] * n
        stk, prev = [], 0
        for log in logs:
            tokens = log.split(":")
            if tokens[1] == "start":
                if stk:
                    result[stk[-1]] += int(tokens[2]) - prev
                stk.append(int(tokens[0]))
                prev = int(tokens[2])
            else:
                result[stk.pop()] += int(tokens[2]) - prev + 1
                prev = int(tokens[2]) + 1
        return result  

Python:

class Solution(object):
    def exclusiveTime(self, n, logs):
        """
        :type n: int
        :type logs: List[str]
        :rtype: List[int]
        """
        ans = [0] * n
        stack = []
        for log in logs:
            fid, soe, tmp = log.split(':')
            fid, tmp = int(fid), int(tmp)
            if soe == 'start':
                if stack:
                    topFid, topTmp = stack[-1]
                    ans[topFid] += tmp - topTmp
                stack.append([fid, tmp])
            else:
                ans[stack[-1][0]] += tmp - stack[-1][1] + 1
                stack.pop()
                if stack: stack[-1][1] = tmp + 1
        return ans  

Python:

def exclusiveTime(self, N, logs):
    ans = [0] * N
    stack = []
    prev_time = 0

    for log in logs:
        fn, typ, time = log.split(':')
        fn, time = int(fn), int(time)

        if typ == 'start':
            if stack:
                ans[stack[-1]] += time - prev_time 
            stack.append(fn)
            prev_time = time
        else:
            ans[stack.pop()] += time - prev_time + 1
            prev_time = time + 1

    return ans

Python:  

def exclusiveTime(self, N, logs):
    ans = [0] * N
    #stack = SuperStack()
    stack = []

    for log in logs:
        fn, typ, time = log.split(':')
        fn, time = int(fn), int(time)

        if typ == 'start':
            stack.append(time)
        else:
            delta = time - stack.pop() + 1
            ans[fn] += delta
            #stack.add_across(delta)
            stack = [t+delta for t in stack] #inefficient

    return ans  

C++:

#include <iostream>
#include <vector>
#include <stack>
#include <sstream>
#include <cassert>

using namespace std;

struct Log {
    int id;
    string status;
    int timestamp;
};

class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> times(n, 0);
        stack<Log> st;
        for(string log: logs) {
            stringstream ss(log);
            string temp, temp2, temp3;
            getline(ss, temp, ':');
            getline(ss, temp2, ':');
            getline(ss, temp3, ':');

            Log item = {stoi(temp), temp2, stoi(temp3)};
            if(item.status == "start") {
                st.push(item);
            } else {
                assert(st.top().id == item.id);

                int time_added = item.timestamp - st.top().timestamp + 1;
                times[item.id] += time_added;
                st.pop();

                if(!st.empty()) {
                    assert(st.top().status == "start");
                    times[st.top().id] -= time_added;
                }
            }
        }

        return times;
    }
};

C++:

class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> res(n, 0);
        stack<int> st;
        int preTime = 0;
        for (string log : logs) {
            int found1 = log.find(":");
            int found2 = log.find_last_of(":");
            int idx = stoi(log.substr(0, found1));
            string type = log.substr(found1 + 1, found2 - found1 - 1);
            int time = stoi(log.substr(found2 + 1));
            if (!st.empty()) {
                res[st.top()] += time - preTime;
            }
            preTime = time;
            if (type == "start") st.push(idx);
            else {
                auto t = st.top(); st.pop();
                ++res[t];
                ++preTime;
            }
        }
        return res;
    }
};

C++:  

class Solution {
public:
    vector<int> exclusiveTime(int n, vector<string>& logs) {
        vector<int> res(n, 0);
        stack<int> st;
        int preTime = 0, idx = 0, time = 0;
        char type[10];
        for (string log : logs) {
            sscanf(log.c_str(), "%d:%[^:]:%d", &idx, type, &time);
            if (type[0] == 's') {
                if (!st.empty()) {
                    res[st.top()] += time - preTime;
                }
                st.push(idx);
            } else {
                res[st.top()] += ++time - preTime;
                st.pop();
            }
            preTime = time;
        }
        return res;
    }
};

    

 

All LeetCode Questions List 题目汇总