I want to Calculate the average of three time in hh:mm:ss format . I tried the following code .
public String calculateAverageOfTime()
{
String timeInHHmmss = "08:00:00 08:00:00 08:00:00";
String[] split = timeInHHmmss.split(" ");
long hh = 0,mm = 0,ss = 0;
for (int i = 0; i < split.length; i++)
{
String[] split1 = split[i].split(":");
hh += Long.valueOf(split1[0].trim());
mm += Long.valueOf(split1[1].trim());
ss += Long.valueOf(split1[2].trim());
}
hh = hh / split.length ;
mm = mm / split.length ;
ss = ss / split.length ;
String hms = String.format("%02d:%02d:%02d", hh,mm,ss);
return hms;
}
This code works well. Is there any efficient way to do this ? . Any method available in Java API .
Your current code is not correct, try for example with input 08:00:00 08:00:00 09:30:00. As #StephenC said it, you cannot average times by averaging hour, minute, second components. Times are like base-60 numbers, you know, 60 seconds + 1 = 1:00 instead of 61.
This is correct, calculating the sum of seconds, then taking average and converting back to time:
public static String calculateAverageOfTime(String timeInHHmmss) {
String[] split = timeInHHmmss.split(" ");
long seconds = 0;
for (String timestr : split) {
String[] hhmmss = timestr.split(":");
seconds += Integer.valueOf(hhmmss[0]) * 60 * 60;
seconds += Integer.valueOf(hhmmss[1]) * 60;
seconds += Integer.valueOf(hhmmss[2]);
}
seconds /= split.length;
long hh = seconds / 60 / 60;
long mm = (seconds / 60) % 60;
long ss = seconds % 60;
return String.format("%02d:%02d:%02d", hh,mm,ss);
}
Using Date, SimpleDateFormat or Joda would make the code easier to understand, but I don't think it can be more efficient than this, as this code does strictly what you want to do, which is averaging a base-60 number.
As others have pointed out, beware of the precision loss when averaging.
You might also want to validate that the input string is in the correct format, otherwise the algorithm will break.
There is no single method in the Java API that will solve this problem. But there are some that will help.
What you need to do is to break the problem down into parts:
Convert time strings to numbers.
Average the numbers
Convert the average back to a time string.
The problems of converting a time string to a number and a number to a time string can be solved using the SimpleDateFormat class, or the 3rd-party Joda time class library. Or you could manually extract the hours/minutes/seconds component, convert each to an integer and then compute the number of seconds.
The problem of averaging 3 numbers can be solved with one line of Java ... though you need to beware of the fact that integer division truncates the result.
For the record, you cannot average 3 times by averaging their three components, as you current code tries to do. It doesn't make mathematical sense.
Something like this should work:
String timeInHHmmss = "08:00:00 08:00:00 08:00:00";
String[] split = timeInHHmmss.split(" ");
long sum = 0L;
SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss");
for (int i = 0; i < split.length; i++)
{
sum += sdf.parse(split[i]).getTime();
}
Date avgDate = new Date((sum/split.length));
System.out.println("avg Date is:"+sdf.format(avgDate));
Convert to java.util.Date and then you can do all kinds of operations.
Check Simple Date Format and Date
Related
I was working on a coding challenge where the goal was to find the smallest time difference in a list of Strings. I gave up and looked at a solution which in all makes sense, but I'm confused on the method parseTime. Can anybody please explain this method?
As I see it, the two hours compared are subtracted and then multiplied by 60 (becuase 60 min in one hour?).
From that the minutes are compared and then I get lost.. Can anybody explain this as simple as possible?
public static void main(String[] args) {
String[] strings = {"10:00am", "11:45pm", "5:00am", "12:01am"};
int i = TimeDifference(strings);
System.out.println(i);
}
public static int TimeDifference(String[] strArr) {
int min = Integer.MAX_VALUE;
for (int i=0; i<strArr.length; i++) {
for (int j=0; j<strArr.length; j++) {
if (i != j) {
int time = parseTime(strArr[j], strArr[i]);
if (time < min) min = time;
}
}
}
return min;
}
private static int parseTime(String time1, String time2) {
int time = (parseHour(time2) - parseHour(time1))*60;
time += parseMin(time2) - parseMin(time1);
if (isMorning(time2) != isMorning(time1)) {
time += 12 *60;
} else if (time <= 0 ) {
time += 24 * 60;
}
return time;
}
private static int parseHour(String time) {
int hour = Integer.valueOf(time.split(":")[0]);
return ( hour == 12? 0: hour);
}
private static int parseMin(String time) {
return Integer.valueOf(time.split(":")[1].replaceAll("[^0-9]",""));
}
private static boolean isMorning(String time) {
String post = time.split(":")[1].replaceAll("[^a-z]","");
return post.toLowerCase().equals("am");
}
Try to debug your code in your mind. Please check my comments below:
int minutes = parseTime("10:00am", "11:45pm");
private static int parseTime(String time1, String time2) {
int minutes = (parseHour(time2) - parseHour(time1))*60;//(11 - 10) * 60 = 60 minutes
minutes += parseMin(time2) - parseMin(time1); //60 + 45 - 0 = 105 minutes
if (isMorning(time2) != isMorning(time1)) { // 10:00am is not morning but 11:45pm is morning
minutes += 12 *60; //so we have to add 12 hours(12 * 60 minutes)
} else if (minutes <= 0 ) {
minutes += 24 * 60;
}
return minutes;
}
This is about regular expressions (RegEx). Let's look at 11:45pm as an example, and break up parseTime:
int time = (parseHour(time2) - parseHour(time1))*60;
The first line makes use of parseHour, which makes use of split, which splits the string around matches of the given regular expression, ":", and then parses the string at index 0 as an integer.
The string "11:45pm" is split into a string array {"11", "45pm"}, and the integer 11 is returned.
time += parseMin(time2) - parseMin(time1);
The second line makes use of parseMin, which similarly splits the given string, but parses the string at index 1 instead. However, the string "45pm" cannot be parsed as an integer, because it contains the non-digits "pm", so we use replaceAll to replace any non-digits with empty strings.
replaceAll replaces each substring of this string that matches the given regular expression with the given replacement.
"[^0-9]" matches any single character not present in 0-9, which is all nondigits, so "45pm".replaceAll("[^0-9]", "") returns "45", which is then parsed as an integer.
The rest has been answered by others already.
Yes, the difference in hours are converted to minutes by multiplying by 60. Then using parseMin the difference in minutes are added to it.
Now we need it check if it is AM or PM. Example : if the times were 10:30 am and 1:35 pm. The actual time difference is +115 minutes.
After parseHour the time would have a value of -600 ((1-10)*60). Then after parseMin, time would be -595 (difference of 5 minutes added). Now we need to see if it is AM or PM. Since time1 is AM and time2 is PM, we add a difference of 12hrs. This is because the same time in AM and PM would differ by 12hrs.
After this the time would be (-605 + 720) = 115 minutes.
The case when time<=0. This is when time2 is before time1. example time1 be 10:35am and time2 be 10:30am. So the question is - how much time would it take to REACH 10:30am FROM 10:35am. Since it is alreaddy 10:35am and since we cannot go back in time, we need to wait till the next day(24 hrs) to reach 10:30am. So we add 24*60 minutes to the time, which would give -5 + (24*60)
i'm working with a query to bring data acording to a given time. so i want to grab a timestamp an remove the last 3 digits of it. Is it possible? how? give me a hint i will totally apreciate that
If you want to REMOVE the digits, divide by 10^n:
long x = 1234567890L;
long removeLastNDigits(long x, long n) {
return x / Math.pow(10, n);
}
removeLastNDigits(x, 3) == 1234567L;
As pointed out by dasblinkenlight, if you want to 0 the last n digits then after removing them you need to add them back as 0:
long x = 1234567890L;
long zeroLastNDigits(long x, long n) {
long tenToTheN = Math.pow(10, n);
return (x / tenToTheN) * tenToTheN;
}
zeroLastNDigits(x, 3) == 1234567000L;
If you want to remove the last 3 digits, just divide by 1000 :
long l = ...;
l = l / 1000;
If you want to change the last 3 digits to 0, divide by 1000 and then multiple by 1000 :
long l = ...;
l = (l / 1000) * 1000;
just divide by 1000.
long x = 123456789L;
long withoutLast3 = x / 1000; // equals to 123456L
I think you should format your timestamp value according to the required format in the database, if that is what you are looking for that is fetching data based on date in a certain format.
You can format your date according based on a dateformatter and then use it in query.
Check
DateFormat
SimpleDateFormat
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I will start with an example: time is 20:00 (2000) and I want to find how many minutes its been since 09:00 (0900), I tried using mod of 60 but this gave me an off time.
Example:
Elapsed time between 1610 1700
1700-1610 = 90, this is obviously wrong
(1700-1610)%60 = 30,
90-30 equals 60, but the answer should be 50. I'm very confused with what I should be doing, how would I go about dealing with this in java? Thanks!
Convert hours to minutes.
1700 hours = 17*60 minutes
1610 hours = 16*60 minutes + 10 minutes
to find out the difference, simple subtraction will do the job
17*60 - 16*60 - 10
Update:
Assuming user enters in 0000 (hhmm) format, you can simply split by size
//psuedo code
String data = userInput;
int hours = Integer.parseInt(data.split(0,2));
int mins = Intger.parseInt(data.split(2,4));
You mixing two representations of the time
13:50
1350 (minutes)
13:50 is 13 hours and 50 minutes, but 1350 is 1350/60 = 22 hours and 1350%60 = 30 minutes
Before you can add, subtract, multiply or divide time you have to convert it to number. In your case:
Elapsed time between 16:10 17:00
16:10 = 16*60+10 = 970
17:00 = 17*60+00 = 1020
17:00 - 16:10 = 1020 - 970 = 50
In Java you could write:
public int minutesFromTime(String time) {
String [] parts = time.split(":");
int hours = Integer.parseInt(parts[0]);
int minutes = Integer.parseInt(parts[1]);
return hours * 60 + minutes;
}
Then
System.out.println("Difference is "
+ (minutesFromTime("17:00") - minutesFromTime("16:10"))
+ " minutes");
To convert your 24-hour time to minutes you need to determine hours and minutes:
int hours = time / 100;
int minutes = time % 100;
And then calculate minutes since midnight:
int minutesSinceMidnight = (hours * 60) + minutes;
You can subtract the minutesSinceMidnight for both times to arrive at the time difference in minutes.
This simple example should do the job:
public static void main(String[] args) {
String after = "2000";
String before = "0900";
int hours = (parseInt(after) - parseInt(before)) / 100;
int minutes = (parseInt(after) % 100) - (parseInt(before) % 100) % 60;
if (minutes < 0) minutes += 60;
System.out.printf("Minutes passed from %s to %s are %d\n", before, after, hours * 60 + minutes);
}
you are dealing with time (60 min), so you can't directly use the math functions (10 base).
So I suggest that you convert your data to Date and then deal with it.
DateFormat dateFormat = new SimpleDateFormat( "HHmm");
Date date = dateFormat.parse("1610");
Date date2 = dateFormat.parse("1700");
long diff = date2.getTime() - date.getTime();
long diffMinutes = diff / (60 * 1000);
You can't do it like this, you are confusing hour and minute as the same unit in your way to do this.
You should separate hour and minute to get the good answer
for example:
String time1 = "1700";
int hour1 = Integer.parseInt(time1.substring(0,2));
int minute1 = Integer.parseInt(time1.substring(2,4));
From int to String to parse
int timeInt = 1715;
String time1 = ""+timeInt;
int hour1 = Integer.parseInt(time1.substring(0,2));
int minute1 = Integer.parseInt(time1.substring(2,4));
Or staying in int
int timeInt = 1715;
int hour = timeInt / 100;
int min = timeInt - hour*100;
If we assume user will always input in XXXX (hhmm) format you can try using this code:
public static void main(String[] args) {
Scanner userInput = new Scanner(System.in);
System.out.print("Enter first time value: ");
int firstTime = userInput.nextInt();
System.out.print("Enter second time value: ");
int secondTime = userInput.nextInt();
int hoursFirst = (int)(firstTime/100);
int minutesFirst = firstTime%100;
int hoursSecond = (int)(secondTime/100);
int minutesSecond = secondTime%100;
int difference = Math.abs((hoursSecond - hoursFirst - 1) * 60 + (60 - minutesFirst) + minutesSecond);
System.out.println("Difference between " + hoursFirst + ":" + minutesFirst
+ " and " + hoursSecond + ":" + minutesSecond + " is " + difference + " minutes.");
}
Output example:
Enter first time value: 1820
Enter second time value: 1610
Difference between 18:20 and 16:10 is 130 minutes.
So for an assignment we had to write a program that takes two times in military time and shows the difference in hours and minutes between them assuming the first time is the earlier of the two times. We weren't allowed to use if statements as it technically has not be learned. Here's an example of what it'd look like run. In quotes I'll put what is manually entered when it is prompted to.
java MilitaryTime
Please enter first time: "0900"
Please enter second time: "1730"
8 hours 30 minutes (this is the final answer)
I was able to quite easily get this part done with the following code:
class MilitaryTime {
public static void main(String [] args) {
Scanner in = new Scanner(System.in);
System.out.println("Please enter the first time: ");
int FirstTime = in.nextInt();
System.out.println("Please enter the second time: ");
int SecondTime = in.nextInt();
int FirstHour = FirstTime / 100;
int FirstMinute = FirstTime % 100;
int SecondHour = SecondTime / 100;
int SecondMinute = SecondTime % 100;
System.out.println( ( SecondHour - FirstHour ) + " hours " + ( SecondMinute
- FirstMinute ) + " minutes " );
}
}
Now my question is something wasn't assigned (or I wouldn't be here!) is there's another part to this question in the book that says to take that program we just wrote and deal with the case where the first time is later than the second. This has really intrigued me about how this would be done and has really stumped me. Again we aren't allowed to use if statements or this would be easy we basically have all the mathematical functions to work with.
An example would be the first time is now 1730 and the second time is 0900 and so now it returns 15 hours 30 minutes.
I would like to suggest to use org.joda.time.DateTime. There are a lot of date and time functions.
Example :
SimpleDateFormat format = new SimpleDateFormat("dd-MM-yyyy hh:mm");
Date startDate = format.parse("10-05-2013 09:00");
Date endDate = format.parse("11-05-2013 17:30");
DateTime jdStartDate = new DateTime(startDate);
DateTime jdEndDate = new DateTime(endDate);
int years = Years.yearsBetween(jdStartDate, jdEndDate).getYears();
int days = Days.daysBetween(jdStartDate, jdEndDate).getDays();
int months = Months.monthsBetween(jdStartDate, jdEndDate).getMonths();
int hours = Hours.hoursBetween(jdStartDate, jdEndDate).getHours();
int minutes = Minutes.minutesBetween(jdStartDate, jdEndDate).getMinutes();
System.out.println(hours + " hours " + minutes + " minutes");
Your expected program will be as below :
SimpleDateFormat format = new SimpleDateFormat("hhmm");
Dates tartDate = format.parse("0900");
Date endDate = format.parse("1730");
DateTime jdStartDate = new DateTime(startDate);
DateTime jdEndDate = new DateTime(endDate);
int hours = Hours.hoursBetween(jdStartDate, jdEndDate).getHours();
int minutes = Minutes.minutesBetween(jdStartDate, jdEndDate).getMinutes();
minutes = minutes % 60;
System.out.println(hours + " hours " + minutes + " minutes");
Output :
8 hours 30 minutes
Normally, when dealing with time calculations of this nature I would use Joda-Time, but assuming that you don't care about the date component and aren't rolling over the day boundaries, you could simply convert the value to minutes or seconds since midnight...
Basically the problem you have is the simple fact that there are 60 minutes in an hour, this makes doing simple mathematics impossible, you need something which is more common
For example, 0130 is actually 90 minutes since midnight, 1730 is 1050 minutes since midnight, which makes it 16 hours in difference. You can simply subtract the two values to get the difference, then convert that back to hours and minutes...for example...
public class MilTimeDif {
public static void main(String[] args) {
int startTime = 130;
int endTime = 1730;
int startMinutes = minutesSinceMidnight(startTime);
int endMinutes = minutesSinceMidnight(endTime);
System.out.println(startTime + " (" + startMinutes + ")");
System.out.println(endTime + " (" + endMinutes + ")");
int dif = endMinutes - startMinutes;
int hour = dif / 60;
int min = dif % 60;
System.out.println(hour + ":" + min);
}
public static int minutesSinceMidnight(int milTime) {
double time = milTime / 100d;
int hours = (int) Math.floor(time);
int minutes = milTime % 100;
System.out.println(hours + ":" + minutes);
return (hours * 60) + minutes;
}
}
Once you start including the date component or rolling over day boundaries, get Joda-Time out
I would do something like:
System.out.println(Math.abs( SecondHour - FirstHour ) + " hours " + Math.abs( SecondMinute - FirstMinute ) + " minutes " );
The absolute value will give you the difference between the two times as a positive integer.
You could do something like this
//Code like you already have
System.out.println("Please enter the first time: ");
int firstTime = in.nextInt();
System.out.println("Please enter the second time: ");
int secondTime = in.nextInt();
//Now we can continue using the code you already wrote by
//forcing the smaller of the two times into the 'firstTime' variable.
//This forces the problem to be the same situation as you had to start with
if (secondTime < firstTime) {
int temp = firstTime;
firstTime = secondTime;
secondTime = temp;
}
//Continue what you already wrote
There are many other ways but this was something I used for similar problems while learning. Also, note that I changed variable names to follow java naming conventions - variables are lowerCamelCase.
I used 3 classes. Lets go over theory first.
We have two times: A and B.
if AtimeDiff = (B-A).....which can be written -(A-B)
if A>B, then timeDiff = 1440- (A-B) [1440 is total minutes in day]
Thus we need to make timeDiff = 1440 - (A-B) and we need to make 1440 term dissapear when A
Lets make a term X = (A-B+1440) / 1440 (notice "/" is integer division.
'if A
'if A>B then X = 1;
Now look at a new term Y = 1440 * X.
'if A
'if A>B then Y = 1440'.
PROBLEM SOLVED. Now just plug into Java Programs. Note what happens if A=B. Our program will assume we know no time passes if times are exact same time. It assumes that 24 hours have passed. Anyways check out the 3 programs listed below:
Class #1
public class MilitaryTime {
/**
* MilitaryTime A time has a certain number of minutes passed at
* certain time of day.
* #param milTime The time in military format
*/
public MilitaryTime(String milTime){
int hours = Integer.parseInt(milTime.substring(0,2));
int minutes = Integer.parseInt(milTime.substring(2,4));
timeTotalMinutes = hours * 60 + minutes;
}
/**
* Gets total minutes of a Military Time
* #return gets total minutes in day at certain time
*/
public int getMinutes(){
return timeTotalMinutes;
}
private int timeTotalMinutes;
}
Class#2
public class TimeInterval {
/**
A Time Interval is amount of time that has passed between two times
#param timeA first time
#param timeB second time
*/
public TimeInterval(MilitaryTime timeA, MilitaryTime timeB){
// A will be shorthand for timeA and B for timeB
// Notice if A<B timeDifferential = TOTAL_MINUTES_IN_DAY - (A - B)
// Notice if A>B timeDifferential = - (A - B)
// Both will use timeDifferential = TOTAL_MINUTES_IN_DAY - (A - B),
// but we need to make TOTAL_MINUTES_IN_DAY dissapear when needed
//Notice A<B following term "x" is 1 and if A>B then it is 0.
int x = (timeA.getMinutes()-timeB.getMinutes()+TOTAL_MINUTES_IN_DAY)
/TOTAL_MINUTES_IN_DAY;
// Notice if A<B then term "y" is TOTAL_MINUTES_IN_DAY(1440 min)
// and if A<B it is 0
int y = TOTAL_MINUTES_IN_DAY * x;
//yay our TOTAL_MINUTES_IN_DAY dissapears when needed.
int timeDifferential = y - (timeA.getMinutes() - timeB.getMinutes());
hours = timeDifferential / 60;
minutes = timeDifferential % 60;
//Notice that if both hours are equal, 24 hours will be shown.
// I assumed that we would knoe if something start at same time it
// would be "0" hours passed
}
/**
* Gets hours passed between 2 times
* #return hours of time difference
*/
public int getHours(){
return hours;
}
/**
* Gets minutes passed after hours accounted for
* #return minutes remainder left after hours accounted for
*/
public int getMinutes(){
return minutes;
}
private int hours;
private int minutes;
public static final int TOTAL_MINUTES_IN_DAY = 1440;//60minutes in 24 hours
}
Class#3
import java.util.Scanner;
public class MilitaryTimeTester {
public static void main (String[] args){
Scanner in = new Scanner(System.in);
System.out.println("Enter time A: ");
MilitaryTime timeA = new MilitaryTime(in.nextLine());
System.out.println("Enter time B: ");
MilitaryTime timeB = new MilitaryTime(in.nextLine());
TimeInterval intFromA2B = new TimeInterval(timeA,timeB);
System.out.println("Its been "+intFromA2B.getHours()+" hours and "+intFromA2B.getMinutes()+" minutes.");
}
}
import java.util.Scanner;
public class TimeDifference{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// read first time
System.out.println("Please enter the first time: ");
int firstTime = in.nextInt();
// read second time
System.out.println("Please enter the second time: ");
int secondTime = in.nextInt();
in.close();
// if first time is more than second time, then the second time is in
// the next day ( + 24 hours)
if (firstTime > secondTime)
secondTime += 2400;
// first hour & first minutes
int firstHour = firstTime / 100;
int firstMinute = firstTime % 100;
// second hour & second minutes
int secondHour = secondTime / 100;
int secondMinute = secondTime % 100;
// time difference
int hourDiff = secondHour - firstHour;
int minutesDiff = secondMinute - firstMinute;
// adjust negative minutes
if (minutesDiff < 0) {
minutesDiff += 60;
hourDiff--;
}
// print out the result
System.out.println(hourDiff + " hours " + minutesDiff + " minutes ");
}
}
This one is done without using ifs and date thingy. you just need to use integer division "/", integer remainder thing"%", and absolute value and celing. might be able to be simplified but im too lazy at moment. I struggled for hours to figure out and seems nobody else got the answer without using more advanced functions. this problem was in Cay Horstmann's Java book. Chapter 4 in Java 5-6 version of the book "Java Concepts"
import java.util.Scanner;
public class MilitaryTime {
public static void main (String[] args){
//creates input object
Scanner in = new Scanner(System.in);
System.out.println("Enter time A: ");
String timeA = in.next();
System.out.println("Enter time B: ");
String timeB = in.next();
//Gets the hours and minutes of timeA
int aHours = Integer.parseInt(timeA.substring(0,2));
int aMinutes = Integer.parseInt(timeA.substring(2,4));
//Gets the hours and minutes of timeB
int bHours = Integer.parseInt(timeB.substring(0,2));
int bMinutes = Integer.parseInt(timeB.substring(2,4));
//calculates total minutes for each time
int aTotalMinutes = aHours * 60 + aMinutes;
int bTotalMinutes = bHours * 60 + bMinutes;
//timeA>timeB: solution = (1440minutes - (aTotalMinutes - bTotalMinutes))
//timeA<timeB: solution is (bTotalMinutes - aTotalMinutes) or
//-(aTotalMinutes - bTotalMinutes)
//we need 1440 term when timea>timeeB... we use mod and mod remainder
//decider is 1 if timeA>timeB and 0 if opposite.
int decider = ((aTotalMinutes - bTotalMinutes +1440)/1440);
// used 4 Special case when times are equal. this way we get 0
// timeDiffference term when equal and 1 otherwise.
int equalsDecider = (int) Math.abs((aTotalMinutes - bTotalMinutes));
//fullDayMaker is used to add the 1440 term when timeA>timeB
int fullDayMaker = 1440 * decider;
int timeDifference = (equalsDecider)* (fullDayMaker - (aTotalMinutes - bTotalMinutes));
// I convert back to hours and minmutes using modulater
System.out.println(timeDifference/60+" hours and "+timeDifference%60+" minutes");
}
}
java.time
Java 8 and later includes the new java.time framework. See the Tutorial.
The new classes include LocalTime for representing a time-only value without date and without time zone.
Another class is Duration, for representing a span of time as a total number of seconds and nanoseconds. A Duration may be viewed as a number of hours and minutes.
By default the Duration class implements the toString method to generate a String representation of the value using the ISO 8601 format of PnYnMnDTnHnMnS where the P marks the beginning and the T separates the date portion from the time portion. The Duration class can parse as well as generate strings in this standard format.
So, the result in example code below is PT8H30M for eight and a half hours. This format is more sensible than 08:30 which can so easily be confused for a time rather than a duration.
String inputStart = "0900";
String inputStop = "1730";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern ( "HHmm" );;
LocalTime start = formatter.parse ( inputStart , LocalTime :: from );
LocalTime stop = formatter.parse ( inputStop , LocalTime :: from );
Duration duration = Duration.between ( start , stop );
Dump to console.
System.out.println ( "From start: " + start + " to stop: " + stop + " = " + duration );
When run.
From start: 09:00 to stop: 17:30 = PT8H30M
import java.util.*;
class Time
{
static Scanner in=new Scanner(System.in);
public static void main(String[] args)
{
int time1,time2,totalTime;
System.out.println("Enter the first time in military:");
time1=in.nextInt();
System.out.println("Enter the second time in military:");
time2=in.nextInt();
totalTime=time2-time1;
String temp=Integer.toString(totalTime);
char hour=temp.charAt(0);
String min=temp.substring(1,3);
System.out.println(hour+" hours "+min+" minutes");
}
}
I'm trying to convert the time format (hh:mm:ss) to this Special Time Format (hhmmt)?
The Special Time Format (STF) is a 5 digit integer value with the format [hhmmt] where hh are the hours, mm are the minutes and t is the tenth of a minute.
For example, 04:41:05 would convert to 4411.
I'm not sure how to convert the seconds value (05) to a tenth of a minute (1).
Edit:
I've incorporated Adithya's suggestion below to convert seconds to a tenth of a minute, but I'm still stuck.
Here is my current code:
String[] timeInt = time.split(":");
String hours = timeInt[0];
String minutes = timeInt[1];
double seconds = Double.parseDouble(timeInt[2]);
int t = (int) Math.round(seconds/6);
if (t>=10) {
int min = Integer.parseInt(minutes);
// min+=1;
t = 0;
}
String stf="";
stf += hours;
stf += minutes;
stf += String.valueOf(t);
int stf2 = Integer.parseInt(stf);
return stf2;
I'm using a String to store the minutes value but it makes it difficult to increment it since it is a String and not a Integer. But when calculating the "t" (tenth of minute) I have to add 1 to minutes if it exceeds 10. If I were to use parseInt again it will exclude the 0 in front of minutes again.
How can I retain the leading zero and still increment the minutes?
Thanks.
I am guessing, its the value of seconds, compared to the tenth of a minute, which is 6 seconds. So formula for t would be
t= seconds/6 //rounded off to nearest integer
This makes sense as this value is always between 0 and 9, as seconds range from 0 and 59, so its always a single digit
For your example
t = 5/6 = 0.833 = rounded off to 1
Note the 6.0f in the division - this helps you avoid the INT truncation.
string FormatSpecialTime(string time)
{
if (time.Length != 8) throw YourException();
int HH, mm, SS, t;
if (!int.TryParse(time.Substring(0, 2), out HH)) HH = 0;
if (!int.TryParse(time.Substring(0, 2), out mm)) mm = 0;
if (!int.TryParse(time.Substring(0, 2), out SS)) SS = 0;
t = (int) Math.Round(SS / 6.0f);
if (t >= 10)
{
mm++;
t = 0;
}
if (mm >= 60)
{
HH += mm / 60;
mm = mm % 60;
}
return HH.ToString() + (mm > 10 ? mm.ToString() : #"0" + mm) + t;
}