An introduction to arrays, continued

Recall from last time… public static void main ( String args[] ) { //define number of rooms final int N = 100; //define hotel rooms boolean rooms[] = new boolean[ N ]; //define occupied (or not) final boolean occupied = true; //initialize to all unoccupied for (int i=0; i<N; i++) { rooms[i] = !occupied; } … } In the code that follows, we will simply deal with rooms numbered 0..N-1. Also, we’ll define a variable called occupied to make things easier to read.

Recall from last time…  How do we determine if a given room k is occupied?

Recall from last time…  How do we determine if a given room k is occupied? if (rooms[k] == occupied) System.out.println( “room “ + k + “ is occupied.” ); else System.out.println( “room “ + k + “ is not occupied.” );

Recall from last time…  How do we determine if a given room k is occupied? if (rooms[k] == occupied) System.out.println( “room “ + (k+1) + “ is occupied.” ); else System.out.println( “room “ + (k+1) + “ is not occupied.” ); A third alternative to the arrays-starting-at-0-but-people- starting-at-1 problem is to represent them internally as 0..N-1 and externally as 1..N by adding 1. But we’ll stick with 0..N-1 both inside and out in these examples.

Recall from last time…  How can we find an unoccupied room?

Recall from last time…  How can we find an unoccupied room? int where = -1; for (int i=0; i<N && where==-1; i++) { if (rooms[i] == !occupied) { where = i; } if (where != -1) System.out.println( “room “ + where + “ is unoccupied.” ); else System.out.println( “Sorry. No vacancy.” );

How can we determine the percentage of rooms that are occupied?

int count = 0; for (int i=0; i<N; i++) { if (rooms[i] == occupied) ++count; } double pcnt = ((double)count) / N * 100.0; System.out.println( pcnt + “% occupied.” );

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free.  Why is this a better approach?  Can you write this code?

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free.  Why is this a better approach? Because we will spread the guest throughout the hotel And because we will equally use each room.  Can you write this code?

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free. int where = -1; while (where==-1) { int k = r.nextInt( N ); if (rooms[k] == !occupied) where = k; }

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free. int where = -1; while (where==-1) { int k = r.nextInt( N ); if (rooms[k] == !occupied) where = k; } What happens when the hotel is full?

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free. int where = -1; for (int attempt=0; attempt<1000 && where==-1; attempt++) { int k = r.nextInt( N ); if (rooms[k] == !occupied) where = k; }

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free. int where = -1; for (int attempt=0; attempt<1000; attempt++) { int k = r.nextInt( N ); if (rooms[k] == !occupied) { where = k; break; } Similar, but with a break.

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free. int where = -1; for (int attempt=0; attempt<1000 && where==-1; attempt++) { int k = r.nextInt( N ); if (rooms[k] == !occupied) where = k; } But there is still a small chance that the hotel has one room. Yet we were unlucky and didn’t find it. What can we do?

Recall from last time…  Write a piece of code to randomly find a free room. It should determine the free room #. It should yield -1 if no rooms are free. int where = -1; for (int attempt=0; attempt<1000 && where==-1; attempt++) { int k = r.nextInt( N ); if (rooms[k] == !occupied) where = k; } for (int i=0; i<N && where==-1; i++) { if (rooms[i] == !occupied)where = i; } if (where != -1)… Same as before.

Recall from last time…  Write a piece of code to find a block of 3 free rooms.  It should determine the lowest room number of the 3 free rooms (otherwise, -1).  Can you write this code?

Recall from last time…  Write a piece of code to find a block of 3 free rooms. It should determine the lowest room number of the 3 free rooms (otherwise, -1). int where = -1; for (int i=0; i<N && where==-1; i++) { if (rooms[i] == !occupied && rooms[i+1] == !occupied && rooms[i+2] == !occupied ) { where = i; } This works well except for one problem. What happens if the hotel is full except for the last room?

Recall from last time… What happens if the hotel is full except for the last room - array bounds exception! How can we fix it?  Write a piece of code to find a block of 3 free rooms. It should determine the lowest room number of the 3 free rooms (otherwise, -1). int where = -1; for (int i=0; i<N && where==-1; i++) { if (rooms[i] == !occupied && rooms[i+1] == !occupied && rooms[i+2] == !occupied ) { where = i; }

Recall from last time…  Write a piece of code to find a block of 3 free rooms. It should determine the lowest room number of the 3 free rooms (otherwise, -1). int where = -1; for (int i=0; i<N-2 && where==-1; i++) { if (rooms[i] == !occupied && rooms[i+1] == !occupied && rooms[i+2] == !occupied ) { where = i; } What happens if the hotel is full except for the last room - array bounds exception! Problem solved!

Recall from last time…  Use another array for wakeup calls.  Use another array to indicate whether or not a room allows smoking or not.  Can you write this code?

Use another array to indicate whether or not a room allows smoking or not.  What data type should we use?  How many do we need?

Use another array to indicate whether or not a room allows smoking or not.  What data type should we use? boolean  How many do we need? One for each room. Another array!

Use another array to indicate whether or not a room allows smoking or not. //define number of rooms final int N = 100; //define hotel rooms boolean rooms[] = new boolean[ N ]; boolean allowsSmoking[] = new boolean[ N ]; //define occupied (or not) final boolean occupied = true; //initialize to all unoccupied for (int i=0; i<N; i++) { rooms[i] = !occupied; allowsSmoking[i] = false; } //rooms allow smoking for (int i=15; i<=25; i++) { allowsSmoking[i] = true; }

How can we modify this to find a free non-smoking room? //Code to randomly find a free room. //It determines the free room #. //It should yield -1 if no rooms are free. int where = -1; for (int attempt=0; attempt<1000 && where==-1; attempt++) { int k = r.nextInt( N ); if (rooms[k] == !occupied) where = k; } for (int i=0; i<N && where==-1; i++) { if (rooms[i] == !occupied)where = i; } if (where != -1)…

How can we modify this to find a free non-smoking room? //Code to randomly find a free room. //It determines the free room #. //It should yield -1 if no rooms are free. int where = -1; for (int attempt=0; attempt<1000 && where==-1; attempt++) { int k = r.nextInt( N ); if (rooms[k] == !occupied && !allowsSmoking[k]) where = k; } for (int i=0; i<N && where==-1; i++) { if (rooms[i] == !occupied && !allowsSmoking[i]) where = i; } if (where != -1)…

Use another array for wakeup calls.  What data type should we use? Let’s make it an integer like military time. Ex. 700 (0700) is 7AM is 1:30 PM.

Use another array for wakeup calls.  Let’s make it an integer like military time. Ex. 700 (0700) is 7AM is 1:30 PM. //define number of rooms final int N = 100; //define hotel rooms boolean[]rooms= new boolean[ N ]; boolean[]allowsSmoking= new boolean[ N ]; int[]wakeupTime= new int[ N ]; //define occupied (or not) final boolean occupied = true; //initialize to all unoccupied for (int i=0; i<N; i++) { rooms[i] = !occupied; allowsSmoking[i] = false; wakeupTime[i] = -1; //indicating no wakup time yet }