[LeetCode] 636. Exclusive Time of Functions 函数的独家时间
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
给一个n个函数运行记录的log数组,表示非抢占式CPU调用函数的情况,即同时只能运行一个函数。格式为id:start or end:time,求每个函数占用cpu的总时间。
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,然后把当前函数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。
Java:
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:
# 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
[LeetCode] 636. Exclusive Time of Functions 函数的独家时间的更多相关文章
- [LeetCode] Exclusive Time of Functions 函数的独家时间
Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find ...
- [leetcode]636. Exclusive Time of Functions函数独占时间
Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find ...
- Leetcode 之 Exclusive Time of Functions
636. Exclusive Time of Functions 1.Problem Given the running logs of n functions that are executed i ...
- 【LeetCode】636. Exclusive Time of Functions 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 栈 日期 题目地址:https://leetcode ...
- 【leetcode】636. Exclusive Time of Functions
题目如下: 解题思路:本题和括号匹配问题有点像,用栈比较适合.一个元素入栈前,如果自己的状态是“start”,则直接入栈:如果是end则判断和栈顶的元素是否id相同并且状态是“start”,如果满足这 ...
- 636. Exclusive Time of Functions 进程的执行时间
[抄题]: Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU ...
- 636. Exclusive Time of Functions
// TODO: need improve!!! class Log { public: int id; bool start; int timestamp; int comp; // compasa ...
- [Swift]LeetCode636. 函数的独占时间 | Exclusive Time of Functions
Given the running logs of n functions that are executed in a nonpreemptive single threaded CPU, find ...
- Java实现 LeetCode 636 函数的独占时间(栈)
636. 函数的独占时间 给出一个非抢占单线程CPU的 n 个函数运行日志,找到函数的独占时间. 每个函数都有一个唯一的 Id,从 0 到 n-1,函数可能会递归调用或者被其他函数调用. 日志是具有以 ...
随机推荐
- linux下安装cryptography兼论查找合适pip的whl文件技巧
cryptography这个包,如果源码安装,需要GCC之类的编译,在生产环境不太现实. 所以选择了whl文件安装. 但在官方提供的whl文件里,没有我们熟悉的cp36-cp36m这样的命名文件,肿么 ...
- docker学习3-镜像的基本使用
前言 Docker的三大核心概念:镜像.容器.仓库.初学者对镜像和容器往往分不清楚,学过面向对象的应该知道类和实例,这跟面向对象里面的概念很相似 我们可以把镜像看作类,把容器看作类实例化后的对象. d ...
- webpack在nodejs中应用(支持es6语法及热加载)
安装 npm i webpack webpack-cli @babel/core babel-loader @babel/preset-env @babel/node clean-webpack-pl ...
- spring-cloud(一)
1.SpringCloud概述和搭建Eureka服务注册中心 Spring Cloud是一系列框架的有序集合.它利用Spring Boot的开发便利性巧妙地简化了分布式系统基础设施的开发,如服务发现注 ...
- 最近公司遇到了APR攻击,顺便了解一下知识
原因及背景 最近公司遇到了APR攻击导致整个公司研发部.测试部.客服部.工程部等几个部门统一无法上网,TV(team viewer)无法使用,部署在公网的B/S架构系统系统无法访问,开发代码上传和下载 ...
- mysql 日期处理
mysql> select curdate(); +------------+ | curdate() | +------------+ | -- | +------------+ row in ...
- 【DataStage】使用Sequence Job报错:CopyOfseq_ld..JobControl (fatal error from @Coordinator): Sequence job (restartable) will abort due to previous unrecoverable errors
错误描述: 在使用Sequence Job加载作业的时候,报了个错,详细错误内容如下: 出现这个错误的原因是由于以下配置问题,Excution action的状态为Run造成. 解决方案: 将Excu ...
- <英狼>--团队作业3 王者光耀--终极版
队员 陶俊宇_031702113 卞永亨_031702229 唐怡_031702109 Github 吉哈---King-Shines 队员输出百分比,数据为估值仅供参考 MVP:队长:陶俊宇 60% ...
- EasyExcel写入百万级数据到多sheet---非注解方式
EasyExcel是什么? 快速.简单避免OOM的java处理Excel工具 一.项目需求 从mongo库中查询数据,导出到excel文件中.但是动态导出的excel有多少列.列名是什么.有多少she ...
- node.js使用cluster实现多进程
首先郑重声明: nodeJS 是一门单线程!异步!非阻塞语言! nodeJS 是一门单线程!异步!非阻塞语言! nodeJS 是一门单线程!异步!非阻塞语言! 重要的事情说3遍. 因为nodeJS天生 ...