Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5] Example 2:
Input: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] Output: [1,2,3,4,8,12,11,10,9,5,6,7]
没有更简便的方法，就是暴力解，O（m*n），方法就是四个变量x1,x2,y1,y2代表四个边界，每次走过一个方向就缩小一个边界，然后判断是否边界越界了：x1>x2或者y1>y2，如果符合则直接break，否则继续循环。
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if (matrix.length == 0 || matrix[0].length == 0) return res;
int x1 = 0;
int x2 = matrix.length-1;
int y1 = 0;
int y2 = matrix[0].length-1;
int x = 0;
int y = 0;
while (x1 <= x2 && y1 <= y2) {
while (y <= y2) res.add(matrix[x][y++]);
x1++;
x++;
y--;
if (x1 > x2) break;
while (x <= x2) res.add(matrix[x++][y]);
y2--;
x--;
y--;
if (y1 > y2) break;
while (y >= y1) res.add(matrix[x][y--]);
x2--;
y++;
x--;
if (x1 > x2) break;
while (x >= x1) res.add(matrix[x--][y]);
y1++;
x++;
y++;
}
return res;
}
