題目如下：

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].

螺旋形狀需要有四個方向，右，下，左，上，再依次進行。當在向右移動時，遇到邊界或者後面一個已經列印過，則改變方向，向下；其它方向依次相同。因此需要有一個二維陣列記錄當前的位置是否列印過，還需要有一個變數記錄已經列印了多少個，當列印的個數和二維陣列總個數相同時，則停止列印。

程式碼如下：

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> list = new ArrayList<Integer>();
        int nRows = matrix.length;
        if(nRows == 0) return list;
        int nCols = matrix[0].length;
        int total_Len = nRows * nCols;//陣列總長度
        int[][] used = new int[nRows][nCols];//用於判斷matrix陣列中的元素是否列印過
        
        int current_DIR = 0;//當前方向
        int i = 0; int j = 0;
        int len = 0;
        while(true){
        	if(len == total_Len){//如果當前列印長度和總長度相等，則停止列印
        		break;
        	}
        	switch(current_DIR){
        	case 0:{//0代表向右的方向
        		if(j == nCols || used[i][j] == 1){//改變方向的邊界條件
        			j--;
        			i++;
        			current_DIR = 1;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			j++;
        		}
        		break;
        	}
        	case 1:{//1代表向下
        		if(i == nRows || used[i][j] == 1){
        			i--;
        			j--;
        			current_DIR = 2;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			i++;
        		}
        		break;
        	}
        	case 2:{//2代表向左
        		if(j == -1 || used[i][j] == 1){
        			j++;
        			i--;
        			current_DIR = 3;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			j--;
        		}
        		break;
        	}
        	case 3:{//3代表向上
        		if(i == -1 || used[i][j] == 1){
        			i++;
        			j++;
        			current_DIR = 0;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			i--;
        		}
        	}
        	}
        }
        return list;
    }
}
 

Spiral Matrix II

Given an integer n, generate a square matrix filled with elements from 1 to n² in spiral order.

For example,
Given n = 3,

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

public class Solution {
    public int[][] generateMatrix(int n) {
         if(n == 0) return new int[0][0];
        int[][] matrix = new int[n][n];
        int len = 0;
        int total_Len = n * n;
        int current_Dir = 0;
        int[][] used = new int[n][n];
        int i = 0, j = 0;
        while(true){
        	if(len == total_Len){
        		break;
        	}
        	switch(current_Dir){
        	case 0:{
        		if(j == n || used[i][j] == 1){
        			j--;
        			i++;
        			current_Dir = 1;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			j++;
        		}
        		break;
        	}
        	case 1:{
        		if(i == n || used[i][j] == 1){
        			i--;
        			j--;
        			current_Dir = 2;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			i++;
        		}
        		break;
        	}
        	case 2:{
        		if(j == -1 || used[i][j] == 1){
        			j++;
        			i--;
        			current_Dir = 3;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			j--;
        		}
        		break;
        	}
        	case 3:{
        		if(i == -1 || used[i][j] == 1){
        			i++;
        			j++;
        			current_Dir = 0;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			i--;
        		}
        		break;
        	}
        	}
        	
        }
        return matrix;
    }
}