如何使这种方法更有效？

Question

So I'm writing a method called getThanksgiving . It works as is and it's part of a much larger class but I needed advice on how to make it more efficient. The getWeekDay method just returns what day of the week November 1 is on a user-inputted year.

public String getThanksgiving(){
String a = getWeekDay(11, 1);
int offset = 0;

    if(a.equals("Friday")){
    offset = 7;
    }

    if(a.equals("Saturday")){
    offset = 6;
    }

    if(a.equals("Sunday")){
    offset = 5;
    }

    if(a.equals("Monday")){
    offset = 4;
    }

    if(a.equals("Tuesday")){
    offset = 3;
    }

    if(a.equals("Wednesday")){
    offset = 2;
    }

    if(a.equals("Thursday")){
    offset = 1;
    }   

 int date = 21 + offset;
 thanksgiving = "Thursday, November " + date; 

 return thanksgiving;
}

I tried rewriting it as a for loop but it's not working.

public String getThanksgiving(){
String a = getWeekDay(11, 1);
int offset = 8;

String[] wTable = {"Friday", "Saturday", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday"};
    for(int i = 1; i < 8; i++){
        if(a.equals(wTable[i - 1])){
        offset --; 
        }
    }
 }

Also, the idea of offset and adding 21 is just something my teacher wants us to do. Thanks in advance!

Answer 1

you can use switch case

like

switch(a )

{
   case "Monday":
    offset = 4;
    break;

  case "Tuesday":
    offset = 3;
    break;

}

reference

switch(n)
{
case 1:
  execute code block 1
  break;
case 2:
  execute code block 2
  break;
default:
  code to be executed if n is different from case 1 and 2
}

Answer 2

I don't think you can make it "more efficient" (ie runtime performance). If you want to make your code more readable shorter, I think you're almost there

String[] wTable = {null, "Thursday", "Wednesday", "Tuesday", "Monday", "Sunday", "Saturday", "Friday"};
for(int i = 1, n = wTable.lenght; i < n; i++) {
    if(a.equals(wTable[i])){
        offset = i;
        break; 
    }
}

Answer 3

To specifically address your question on 'how to make this method more efficient', one thing to note is that the method evaluates every if statement even in the case when you already found your solution. Using 'bare bones' java if's you could add a condition to check when the day has been found in this way:

int offset = 0;
boolean found = false;
if(!found && a.equals("Friday")){
    offset = 7;
    found = true;
}

if(!found && a.equals("Saturday")){
    offset = 6;
    found = true;
}

This flag will marginally reduce runtime by virtue of shortcut evaluation of the && (and) operator, only executing the string comparison until a match is found. You can achieve a similar performance result with a for and using break to get out of the loop when you have found a matching element.

A better alternative would be using a Map data structure:

Map<String, Integer> daysByOffset = new HashMap<String,Integer>();
// this 'setup' part you only do once
daysByOffset.put("Friday", 7);
daysByOffset.put("Saturday", 6);
...

Then the lookup part is very efficient, as the lookup in a hashmap is O(1):

int offset = daysByOffset.get(day);

An elegant alternative would be using enums that encapsulate the offset information:

public enum DaysWithOffset {
    FRIDAY(7), SATURDAY(6),..., THURSDAY(1);
    private final offset;
    private DaysWithOffset(int offset) {
        this.offset = offset;
    }

    public int getOffset() {
        return offset;
    }
}

After the enum is defined, each enum constant will contain the corresponding offset info:

FRIDAY.getOffset() // = 7

You can calculate the offset by resolving the enum from the provided String and asking the offset value from it:

...
String a = getWeekDay(11, 1);
int offset = DaysWithOffset.valueOf(day.toUpperCase()).getOffset();
...

Coming back to the question on which option is more efficient, both map and enum have an O(1) lookup, (the Enum lookup being slightly better by the optimized internal enum dictionary. Yet, the enum operation requires a toUpperCase(), while the map doesn't) both these options will perform a lot better than a list of if's (the original version) or a for loop.

I include these options for completeness of the answer and also for you to have a 'sneak preview' of the possibilities that the Java language offers.

Now a teaser: If you were doing this in Scala, you'd write something like this: val daysByOffset = Map("Friday" -> 7, "Saturday" -> 6,...,"Thursday" ->1) def thanksGiving(day:String):String = "Thursday, November " + (daysByOffset(day)+21)

If I were learning a language today, it would be Scala.

Answer 4

how about this?

private final static Map<String, int> dayMap = new HashMap<String,int>() 
{
dayMap.put("Monday", 0);
// do for the rest
};

in your method:

public String getThanksgiving(){
    String a = getWeekDay(11, 1);
    //do a lookup
    int result = dayMap.get(a);
    // do somthing with it. and return
    return "blah "+ result;
}