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Write a routine that prints out a 2-D array in spiral order.
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Write a routine that prints out a 2-D array in spiral order.
5,878 Views | (11 votes, avg: 4.45)
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I don’t think I could write the correct code for this question in 15 minutes. The code below only show clockwise printing result. You are welcomed to optimize or generalize it.
#include <stdio.h>
char test[5][5] = { “ABCD”, “ABCD”, “ABCD”, “ABCD” };
void main() {
int x_count = 4, x_limit = 4, y_count = 4, y_limit = 4;
int pivol = 0, // x changed = 0, y changed = 1
i = 0,
j = 0,
prev_i = 1,
prev_j = 1,
temp;
while (x_limit != 0 && y_limit != 0) {
if (pivol == 0) {
if (x_count == x_limit) {
putchar(test[i][j]);
pivol = 1;
y_count = 1;
y_limit–;
temp = j;
j = prev_j;
prev_j = temp;
continue;
}
if (prev_i <= i) {
putchar(test[i][j]);
prev_i = i; i++; x_count++;
continue;
}
if (prev_i > i) {
putchar(test[i][j]);
prev_i = i; i–; x_count++;
}
}
if (pivol == 1) {
if (y_count == y_limit) {
putchar(test[i][j]);
pivol = 0;
x_count = 1;
x_limit–;
temp = i;
i = prev_i;
prev_i = temp;
continue;
}
if (prev_j <= j) {
putchar(test[i][j]);
prev_j = j; j++; y_count++;
continue;
}
if (prev_j > j) {
putchar(test[i][j]);
prev_j = j; j–; y_count++;
}
}
}
}
What do you mean by printing array in spiral fashion??
If you have 2D array as
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
Does spiral printing mean ” 1 2 2 3 3 3 4 4 4… and so on? or printing 12345 23456 3456….
To ssanand:
You may probe what the interview intends to ask you.
Hi,
This is what I think is Spiral Traversal -
If you were given an array like:
A B C D E
1 2 3 4 5
A B C D E
1 2 3 4 5
The output should be: ABCDE5E54321A1234DCB
Key: All the elements of the array should be covered.
Can you modify the code accordingly zanic?
Thx!
This program works !!!
I have hard-coded a 4×4 matrix called arr
Run the program and see the results..
Anytime you want to use a different size matrix.. just change the row and col values that I initialized below..works for all cases that I tested..
int main ()
{
int arr[][4] = { {1,2,3,4}, {5,6,7,8}, {9,10,11,12}, {13, 14, 15, 16}};
int i = 0, j = 0, row = 4, col = 3, done_i = 0, done_j = 0;
int count = row*col;
for (; i < row-done_i; i++)
{
for (; j < col-done_j; j++)
{
cout << arr[i][j] << ” “;
cout.flush(); count–;
}
for (j –, i++; i < row-done_i; i++)
{
cout << arr[i][j] << ” “;
cout.flush(); count–;
}
done_i++;
if (count <= 0)
break;
for (i–, j–; j >= done_j; j–)
{
cout << arr[i][j] << ” “;
cout.flush(); count–;
}
for (j++, i–; i >= done_i; i–)
{
cout << arr[i][j] << ” “;
cout.flush(); count–;
}
done_j++;
j++;
}
return 1;
}
Here is my universal solution. You can just copy and paste to run:
static void spiralRoute()
{
char a[7][7] = {
{’a', ‘b’, ‘c’, ‘d’, ‘e’, ‘f’, ‘g’},
{’h', ‘i’, ‘j’, ‘k’, ‘l’, ‘m’, ‘n’},
{’o', ‘p’, ‘q’, ‘r’, ’s’, ‘t’, ‘u’},
{’1′, ‘2′, ‘3′, ‘4′, ‘5′, ‘6′, ‘7′},
{’O', ‘P’, ‘Q’, ‘R’, ‘S’, ‘T’, ‘U’},
{’H', ‘I’, ‘J’, ‘K’, ‘L’, ‘M’, ‘N’},
{’A', ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’},
};
char rightoutput[50] = “abcdefgnu7UNGFEDCBAHO1ohijklmt6TMLKJIP2pqrs5SRQ34″;
char myoutput[50];
/* DIRECTIONS:
0: towarding right;
1: towarding down;
2: towarding left;
3: towarding up;
*/
int direction = 0;
int up_ = 0, down_ = 6, right_ = 6, left_ = 0;
int i, counter = 0, bOver = 0;
while (!bOver) {
switch (direction)
{
case 0:
if (right_ < left_)
{
bOver = 1;
} else {
for (i = left_; i <= right_; i++)
{
myoutput[counter++] = a[up_][i];
}
up_++;
direction++;
}
break;
case 1:
if (down_ < up_)
{
bOver = 1;
} else {
for (i = up_; i <= down_; i++)
{
myoutput[counter++] = a[i][right_];
}
right_–;
direction++;
}
break;
case 2:
if (right_ < left_)
{
bOver = 1;
} else {
for (i = right_; i >= left_; i–)
{
myoutput[counter++] = a[down_][i];
}
down_–;
direction++;
}
break;
case 3:
if (down_ < up_)
{
bOver = 1;
} else {
for (i = down_; i >= up_; i–)
{
myoutput[counter++] = a[i][left_];
}
left_++;
direction = 0;
}
break;
}
}
myoutput[counter] = ‘ ’;
printf(”rightoutput = %sn”, rightoutput);
printf(”myoutput %sn”, myoutput);
if (strcmp(rightoutput, myoutput) == 0) {
printf(”passedn”);
} else {
printf(”not passedn”);
}
}
void print_spiral(int** arr, int size)
{
int i,j,k,middle;
for(i=size-1, j=0; i>0; i–, j++) {
for(k=j; k<i; k++)
printf(”%d”, arr[j][k]);
for(k=j; k<i; k++)
printf(”%d”, arr[k][i]);
for(k=i; k>j; k–)
printf(”%d”, arr[i][k]);
for(k=i; k>j; k–)
printf(”%d”, arr[k][j]);
}
middle = (n-1)/2;
if (n % 2 == 1)
printf(”%d”, arr[middle][middle]);
printf(”n”);
}
# include
int arr[4][4] = {
{1,2,3,4},
{5,6,7,8},
{9,10,11,12},
{13, 14, 15, 16}
};
typedef unsigned int uint;
void main()
{
for( uint size=4, idx=0; size>0; size-=2, ++idx )
{
for( uint i=0; i0; –ii ) cout 1; –jj ) cout
Algo for spiral traversal can be as follows
int Switch = 1;
int max_row = MAXROW;
int max_col = MAXCOL;
int i = j = 0;
while ( max_col > 0 or max_row > 0 )
{
for( k = 0 ; k
void printShape(char **bas,int x, int y)
{
int k =0;
while(x>1 && y>1)
{
//Sag
for(int i=k; ik; i–)
printf(”%c “,bas[y-1+k][i]);
//Yukari
for(int i=y-1+k; i>k; i–)
printf(”%c “,bas[i][k]);
x= x - 2;
y= y - 2;
k++;
}
if(x==1 && y==1){
printf(”%c “,bas[0][0]);
}
return;
}
void printShape(char **bas,int x, int y)
{
int k =0;
while(x>1 && y>1)
{
//Sag
for(int i=k; ik; i–)
printf(”%c “,bas[y-1+k][i]);
//Yukari
for(int i=y-1+k; i>k; i–)
printf(”%c “,bas[i][k]);
x= x - 2;
y= y - 2;
k++;
}
if(x==1 && y==1){
printf(”%c “,bas[0][0]);
}
return;
}
in VB 6.0：
Private Function Min(ByVal a As Integer, ByVal b As Integer)
If a
In Java, the correct solution for above is like this it can work for n by n matix or 2d array. Just replace with correct values:
public class PrintCircular
{
public static void main(String args[])
{
final int right = 1;
final int down = -1;
final int left = 2;
final int up = -2;
int arr[][] = { {1,2,3,4,5,6}, {7,8,9,10,11,12}, {13, 14, 15, 16, 17,18}, {19,20,21,22,23,24}, {25,26,27,28,29,30},
{31,32,33,34,35,36}};
int count = 36;
int i = 0, j=0, minx = 0, maxx = 5, miny = 0, maxy = 5;
int direction = right;
while(count>0)
{
System.out.print(arr[i][j] + ” “);
if(direction == right)
j++;
else if(direction == down)
i++;
else if(direction == left)
j–;
else
i–;
if(j==miny && i==minx)
direction = right;
else if(j==maxy && i==minx)
direction = down;
else if(i==maxx && j==maxy)
direction = left;
else if(i==maxx && j==miny)
direction = up;
if(i==minx && j==miny)
{
minx++;
maxy–;
miny++;
maxx–;
i=minx;
j=miny;
}
System.out.print(”minx =” + minx + ” maxx = ” + maxx + ” miny = ” + miny + ” maxy = ” + maxy + ” i = ” + i
+ ” j =” +j + “\n”);
count –;
}
}
}
(C++and recursive, based on #8)
template
void print_spiral(T** arr,int start,int row,int col)
{
if((start right
for(int j = start ; j
for(int i = start+1; i left
for(int j = col -1 ; j > start ; j–)
cout top
for(int i = row -1 ;i > start; i–)
cout bottom
for(int i = start; i
function TravMatrix( int matrix[][], int row , int col )
{
variable int rowU = 0, rowD = 0, colL = 0,colR = 0,total = 0,r=0,c=0;
1. total = rowU + rowD + colL + colR ;
2. while total != ( row + col )
do
1. if rowTraverseRight = true then
1. for ( ; c
do
2. print matrix[r][c];
done
3. rowD++;
4. colTraverseUp = true;
5. rowTraverseRight = colTraverseDown = rowTraverseLeft = false;
4. else if colTraverseUp = true then
1. for ( ; r > rowD ; r– )
do
2. print matrix[r][c];
done
3. colL++;
4. rowTraverseRight = true;
5. colTraverseDown = rowTraverseLeft = colTraverseUp = false;
5. total = rowU + rowD + colL + colR ;
done
3. Exit From Function.
}
enum DIRECTION {
DOWN, RIGHT, UP, LEFT
}
public static void arrayAsSpiral(Object[][] P_array) {
int x_size = P_array.length;
int y_size = 0;
for (Object[] obj : P_array) {
if (obj.length > y_size) {
y_size = obj.length;
}
}
int x_min = 0;
int x_max = x_size - 1;
int y_min = 0;
int y_max = y_size - 1;
int p_x = 0;
int p_y = 0;
DIRECTION dir = DIRECTION.DOWN;
System.out.print(”[” + P_array[p_x][p_y] + “]”);
while ((y_max >= y_min) && (x_max >= x_min)) {
switch (dir) {
case DOWN:
if (p_y y_min) {
p_y–;
} else {
dir = DIRECTION.LEFT;
x_max–;
continue;
}
break;
case LEFT:
if (p_x > x_min) {
p_x–;
} else {
dir = DIRECTION.DOWN;
y_min++;
continue;
}
break;
}
System.out.print(”[” + P_array[p_x][p_y] + “]”);
}
}
omkar
dot net code
public static void zigzag(int[,] array,int maxRow,int maxCol)
{
int row = 0,col=0;
bool flag = false;
for (int i = 0; i = 0; j–)
{
row = i - j;
col = j;
if (row
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