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SPIRAL ARRAY IN CLOCKWISE
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specification:- | |
Given a matrix of m * n elements (m rows, n columns), return all elements of the matrix in spiral order. | |
Example:- | |
Input | |
[ | |
[ 1, 2, 3 ], | |
[ 4, 5, 6 ], | |
[ 7, 8, 9 ] | |
] | |
output | |
[1, 2, 3, 6, 9, 8, 7, 4, 5] | |
Explanation- | |
top-> 1 2 3 4 | |
6 8 9 6 | |
btm-> 2 4 3 5 | |
^ ^ | |
| | | |
lft rht | |
########################JAVA CODE###################################################################### | |
public class Solution { | |
// DO NOT MODIFY THE ARGUMENTS WITH "final" PREFIX. IT IS READ ONLY | |
public int[] spiralOrder(final int[][] mat) { | |
int k=0; | |
int m=mat.length; | |
int n=mat[0].length; | |
if(m==1 && n==1) | |
return mat[0]; | |
int arr[]=new int[m*n]; | |
int top=0,btm=m-1,lft=0,rht=n-1,dir=0; | |
while(top<=btm && lft<=rht) | |
{ | |
if (dir==0) //Move left to right | |
{ | |
for(int i=top;i<=rht;i++) | |
{ | |
arr[k++]=mat[top][i]; | |
} | |
top=top+1; | |
dir=1; | |
} | |
else if(dir==1)//Move top to bottom | |
{ | |
for(int i=top;i<=btm;i++) | |
{ | |
arr[k++]=mat[i][rht]; | |
} | |
rht=rht-1; | |
dir=2; | |
} | |
else if(dir==2)//move right to left | |
{ | |
for(int i=rht;i>=lft;i--) | |
{ | |
arr[k++]=mat[btm][i]; | |
} | |
btm=btm-1; | |
dir=3; | |
} | |
else if(dir==3) ///move bottom to top | |
{ | |
for(int i=btm;i>=top;i--) | |
{ | |
arr[k++]=mat[i][lft]; | |
} | |
lft=lft+1; | |
dir=0; | |
} | |
} | |
return arr; | |
} | |
} |
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