LeetCode – Spiral Matrix (Java)

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example, given the following matrix:

[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

You should return [1,2,3,6,9,8,7,4,5].

Java Solution 1

If more than one row and column left, it can form a circle and we process the circle. Otherwise, if only one row or column left, we process that column or row ONLY.

public class Solution {
    public ArrayList<Integer> spiralOrder(int[][] matrix) {
        ArrayList<Integer> result = new ArrayList<Integer>();
 
        if(matrix == null || matrix.length == 0) return result;
 
        int m = matrix.length;
        int n = matrix[0].length;
 
        int x=0; 
        int y=0;
 
        while(m>0 && n>0){
 
            //if one row/column left, no circle can be formed
            if(m==1){
                for(int i=0; i<n; i++){
                    result.add(matrix[x][y++]);
                }
                break;
            }else if(n==1){
                for(int i=0; i<m; i++){
                    result.add(matrix[x++][y]);
                }
                break;
            }
 
            //below, process a circle
 
            //top - move right
            for(int i=0;i<n-1;i++){
                result.add(matrix[x][y++]);
            }
 
            //right - move down
            for(int i=0;i<m-1;i++){
                result.add(matrix[x++][y]);
            }
 
            //bottom - move left
            for(int i=0;i<n-1;i++){
                result.add(matrix[x][y--]);
            }
 
            //left - move up
            for(int i=0;i<m-1;i++){
                result.add(matrix[x--][y]);
            }
 
            x++;
            y++;
            m=m-2;
            n=n-2;
        }
 
        return result;
    }
}

Java Solution 2

We can also recursively solve this problem. The solution’s performance is not better than Solution or as clear as Solution 1. Therefore, Solution 1 should be preferred.

public class Solution {
    public ArrayList<Integer> spiralOrder(int[][] matrix) {
        if(matrix==null || matrix.length==0) 
            return new ArrayList<Integer>();
 
        return spiralOrder(matrix,0,0,matrix.length,matrix[0].length);
    }
 
 
    public ArrayList<Integer> spiralOrder(int [][] matrix, int x, int y, int m, int n){
        ArrayList<Integer> result = new ArrayList<Integer>();
 
        if(m<=0||n<=0) 
            return result;
 
        //only one element left
        if(m==1&&n==1) {
            result.add(matrix[x][y]);
            return result;
        }
 
        //top - move right
        for(int i=0;i<n-1;i++){
            result.add(matrix[x][y++]);
        }
 
        //right - move down
        for(int i=0;i<m-1;i++){
            result.add(matrix[x++][y]);
        }
 
        //bottom - move left
        if(m>1){    
            for(int i=0;i<n-1;i++){
                result.add(matrix[x][y--]);
            }
        }
 
        //left - move up
        if(n>1){
            for(int i=0;i<m-1;i++){
                result.add(matrix[x--][y]);
            }
        }
 
        if(m==1||n==1)
            result.addAll(spiralOrder(matrix, x, y, 1, 1));
        else    
            result.addAll(spiralOrder(matrix, x+1, y+1, m-2, n-2));
 
        return result;
    }
}