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, 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.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
不知道具体怎么写,感觉stack的题目前就是背 暂时没有发现规律
[一句话思路]:
更新数组、更新preTime、更新stack
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
感觉stack的题目前就是背 暂时没有发现规律
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
涉及到优先顺序的存储结构,一般都是stack
[算法思想:递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
class Solution {
public int[] exclusiveTime(int n, List<String> logs) {
//ini: res, split, String[3], prevTime, stack
Stack<Integer> stack = new Stack<Integer>();
int[] res = new int[n];
int prevTime = 0; //cc
if (logs == null || logs.size() == 0) return res; //for loop:
for (String log : logs) {
String[] words = log.split(":");
//update res
if (!stack.isEmpty()) res[stack.peek()] += Integer.parseInt(words[2]) - prevTime;
//update prevTime
prevTime = Integer.parseInt(words[2]);
//update stack
if (words[1].equals("start")) stack.push(Integer.parseInt(words[0]));
else {
res[stack.pop()]++;
prevTime++;
}
} return res;
}
}
636. Exclusive Time of Functions 进程的执行时间的更多相关文章
- 【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函数独占时间
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
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
// TODO: need improve!!! class Log { public: int id; bool start; int timestamp; int comp; // compasa ...
- Leetcode 之 Exclusive Time of Functions
636. Exclusive Time of Functions 1.Problem Given the running logs of n functions that are executed i ...
- 通过定时任务 bash 脚本 控制 进程 的 执行时间
通过定时任务 bash 脚本 控制 进程 的 执行时间
- [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 ...
- Exclusive Time of Functions
On a single threaded CPU, we execute some functions. Each function has a unique id between 0 and N- ...
随机推荐
- 可以方便配合 Git 的现代编辑器
可以方便配合 Git 的现代编辑器 虽然有些人说编辑器不重要,有人用记事本来编辑 PHP(我是不推荐的),但学是要推荐一些现在编辑器. 可以很方便的配合 Git,比如有冲突时不会看得晕乎乎,使用鼠标点 ...
- ibernate+Struts2环境如何使用jqGrid。
因为公司项目需要,在Hibernate+Struts2的环境下,研究了一下如何使用jqGrid. 说实在的,Struts2+jqGrid不是一个很好的组合.因为jqGrid中很多功能,基本上都使用的是 ...
- volatile浅析
volatile的意思是不稳定的.易变的,但在多线程中却跟字面意思一毛钱关系没有.作为一个变量修饰符,它主要有两个作用:一个是告诉大家,该变量是一个在多个线程之间均可见的变量:另一个是告诉java虚拟 ...
- spring装载配置文件失败报错:org.springframework.beans.factory.xml.XmlBeanDefinitionStoreException
Tomcat容器启动失败,找到 debug日志一看: Context initialization failed org.springframework. beans.factory.xml.XmlB ...
- webpack中imports-loader,exports-loader,expose-loader的区别
Webpack有几个和模块化相关的loader,imports-loader,exports-loader,expose-loader,比较容易混淆.今天,我们来理一理. imports-loader ...
- OpenCV在debug和release模式下选择不同的lib静态库文件
这两天测试OpenCV显示到MFC的Picture控件上,终于测试成功了,但是换到release模式下就会imread失败.发现问题是导入的lib问题. 因为VS如果通过Property Manage ...
- START WITH...CONNECT BY PRIOR详解
START WITH...CONNECT BY PRIOR详解 START WITH...CONNECT BY PRIOR详解 ORACLE中的SELECT语句可以用START WITH...CONN ...
- HTML5的离线存储有几种方式?
localStorage长期存储数据,浏览器关闭后数据不丢失: sessionStorage数据在浏览器关闭后自动删除.
- GIL与线程、进程、协程
GIL全局解释器锁: 1.相信大家都知道python代码是不能直接被机器cpu识别和执行的,它要经过python解释器(也就是我们执行时候的python3 name.py)被编译成机器语言,pytho ...
- 星型打分插件 bootstrap-rating-input
最近帮人实现一个打分的功能,发现bootstrap-rating-input是个简单又好用的星型打分,我对其做了些定制,添加了分值说明,并修改了样式,毕竟 bootstrap 自身的黑色五角星还是不够 ...