lintcode-medium-Spiral Matrix

Posted 2020-07-05 哥布林工程师

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

 

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].

 

public class Solution {
    /**
     * @param matrix a matrix of m x n elements
     * @return an integer list
     */
    public List<Integer> spiralOrder(int[][] matrix) {
        // Write your code here
        
        List<Integer> res = new ArrayList<Integer>();
        
        if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0)
            return res;
        
        int m = matrix.length;
        int n = matrix[0].length;
        int layer = 0;
        
        while((layer < m - layer - 1) && (layer < n - 1 - layer)){
            
            for(int i = 0; i < n - layer - 1 - layer; i++)
                res.add(matrix[layer][layer + i]);
            
            for(int i = 0; i < m - layer - 1 - layer; i++)
                res.add(matrix[layer + i][n - layer - 1]);
            
            for(int i = 0; i < n - layer - 1 - layer; i++)
                res.add(matrix[m - layer - 1][n - layer - 1 - i]);
            
            for(int i = 0; i < m - layer - 1 - layer; i++)
                res.add(matrix[m - layer - 1 - i][layer]);
            
            layer++;
        }
        
        if(m > n){
            if(layer == n - 1 - layer){
                for(int i = layer; i <= m - layer - 1; i++)
                    res.add(matrix[i][layer]);
            }
        }
        else if(m < n){
            if(layer == m - 1 - layer){
                for(int i = layer; i <= n - layer - 1; i++)
                    res.add(matrix[layer][i]);
            }
        }
        else{
            if(layer == m - 1 - layer){
                res.add(matrix[layer][layer]);
            }
        }
        
        return res;
    }
    
}