【Leetcode】Spiral Matrix (Rotated)

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]

.

这道题让螺旋形输出矩阵。先附上代码，再逐块解析

public ArrayList<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> result = new ArrayList<Integer>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return result;
int row = matrix.length;
int col = matrix[0].length;
int x = 0;
int y = 0;
while (row > 0 && col > 0) {
if (row == 1) {
for (int i = 0; i < col; i++)
result.add(matrix[x][y++]);
break;
}
if (col == 1) {
for (int i = 0; i < row; i++)
result.add(matrix[x++][y]);
break;
}

for (int i = 0; i < col - 1; i++)
result.add(matrix[x][y++]);
for (int i = 0; i < row - 1; i++)
result.add(matrix[x++][y]);
for (int i = 0; i < col - 1; i++)
result.add(matrix[x][y--]);
for (int i = 0; i < row - 1; i++)
result.add(matrix[x--][y]);
row = row - 2;
col = col - 2;
x++;
y++;
}
return result;
}

首先是只有一行数据或者一列数据的时候应该怎么处理的问题

显然就直接一行或一列全部添加然后break就OK

if (row == 1) {
for (int i = 0; i < col; i++)
result.add(matrix[x][y++]);
break;
}
if (col == 1) {
for (int i = 0; i < row; i++)
result.add(matrix[x++][y]);
break;
}

然后是核心代码，也就是螺旋形地添加元素 这里一定记着col-1和row-1，还有就是x++，x--，y++，y--一定要区分清楚。

for (int i = 0; i < col - 1; i++)
result.add(matrix[x][y++]);
for (int i = 0; i < row - 1; i++)
result.add(matrix[x++][y]);
for (int i = 0; i < col - 1; i++)
result.add(matrix[x][y--]);
for (int i = 0; i < row - 1; i++)
result.add(matrix[x--][y]);

最后记着更新col和row还有x和y

row = row - 2;
col = col - 2;
x++;
y++;

三大块操作一块都不能少，要非常熟悉。