矩阵螺旋遍历Spiral Matrix，Spiral Matrix2

SpiralMatrix：

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].

算法分析：其实就是两重循环遍历，每次都是遍历一圈，然后遍历里面一圈...，需要记录m,n,每次遍历后，m，n都减去2；还要记录坐标x，y；

特例是n=1；m=1；其实m，n为奇数才有这种特例。

public class SpiralMatrix

{

	public List<Integer> spiralOrder(int[][] matrix)

	{

		List<Integer> res = new ArrayList<>();

		if(matrix.length == 0 || matrix == null)

		{

			return res;

		}

		int m = matrix.length;

		int n = matrix[0].length;//得先判断矩阵是否为空，否则数组下标越界

		int x = 0, y = 0;

		while(m > 0 && n > 0)

		{

			if(m == 1)

			{

				for(int i = 0; i < n; i ++)

				{

					res.add(matrix[x][y++]);

				}

				break;//别忘了break

			}

			else if(n == 1)

			{

				for(int j = 0; j < m; j ++)

				{

					res.add(matrix[x++][y]);

				}

				break;

			}

			for(int i = 0; i < n-1; i ++)

			{

				res.add(matrix[x][y++]);

			}

			for(int i = 0; i < m-1; i ++)

			{

				res.add(matrix[x++][y]);

			}

			for(int i = 0; i < n-1; i ++)

			{

				res.add(matrix[x][y--]);

			}

			for(int i = 0; i < m-1; i ++)

			{

				res.add(matrix[x--][y]);

			}

			x++;

			y++;

			m -= 2;

			n -= 2;

		}

		return res;

	}

}

SpiralMatrix2:

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

For example,
Given n = 3,

You should return the following matrix:

[

 [ 1, 2, 3 ],

 [ 8, 9, 4 ],

 [ 7, 6, 5 ]

]

算法分析：和上一算法，类似，只不过这是个n*n矩阵，其实可以延伸到m*n矩阵。n*n使清空更简单，特别是n=1的时候。

public class SpiralMatrix2

{

	public int[][] generateMatrix(int n)

	{

		int[][] res = new int[n][n];

		if(n == 0)

		{

			return res;

		}

		int x = 0, y = 0;

		int val = 1;

		while(n > 0)

		{

			if(n == 1)

			{

				res[x][y] = val;

				break;

			}

			for(int i = 0; i < n-1; i ++)

			{

				res[x][y++] = val;

				val ++;

			}

			for(int i = 0; i < n-1; i ++)

			{

				res[x++][y] = val;

				val ++;

			}

			for(int i = 0; i < n-1; i ++)

			{

				res[x][y--] = val;

				val ++;

			}

			for(int i = 0; i < n-1; i ++)

			{

				res[x--][y] = val;

				val ++;

			}

			x ++;

			y ++;

			n -= 2;

		}

		return res;

	}

}

巴特西

矩阵螺旋遍历Spiral Matrix，Spiral Matrix2

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