Spiral Traversal of a 2D Matrix
Given a matrix of m x 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, 10, 11, 12},
{13, 14, 15, 16},
{17, 18, 19, 20}
}
Output: [1, 2, 3, 4, 8, 12, 16, 20, 19, 18, 17, 13, 9, 5, 6, 7, 11, 15, 14, 10]
import java.util.ArrayList;
import java.util.List;
public class App {
public static void main(String[] args) {
int[][] mat = {
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16},
{17, 18, 19, 20}
};
System.out.println(spiralOrder(mat));
}
public static List<Integer> spiralOrder(int[][] matrix) {
List<Integer> spiral = new ArrayList<>();
if(matrix == null || matrix.length == 0)
return spiral;
int rowStart = 0;
int rowEnd = matrix.length - 1;
int columnStart = 0;
int columnEnd = matrix[0].length - 1;
while(rowStart <= rowEnd && columnStart<= columnEnd){
for(int i = columnStart; i <= columnEnd; i++){
spiral.add(matrix[rowStart][i]);
}
rowStart++;
for(int i = rowStart; i <= rowEnd; i++){
spiral.add(matrix[i][columnEnd]);
}
columnEnd--;
if(rowStart <= rowEnd){
for(int i = columnEnd; i>= columnStart; i--){
spiral.add(matrix[rowEnd][i]);
}
}
rowEnd--;
if(columnStart <= columnEnd){
for(int i = rowEnd; i >= rowStart; i--){
spiral.add(matrix[i][columnStart]);
}
}
columnStart++;
}
return spiral;
}
}The above implementation have runtime complexity of O(m * n) and space complexity of O(1), where m is number of rows in input matrix and n is the number of columns in the input matrix.
Runtime Complexity = O(m*n)
Space Complexity = O(1)