Description
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].
Solution
Iteration, time O(m*n), space O(1)
注意边界条件即可。
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> list = new LinkedList<>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return list;
}
int m = matrix.length;
int n = matrix[0].length;
for (int i = 0; i < (Math.min(m, n) + 1) / 2; ++i) {
// Traverse right
for (int j = i; j < n - i; ++j) {
list.add(matrix[i][j]);
}
if (i == m - i - 1) { // only one row
continue;
}
// Traverse down
for (int j = i + 1; j < m - i - 1; ++j) {
list.add(matrix[j][n - i - 1]);
}
// Traverse left
for (int j = n - i - 1; j >= i; --j) {
list.add(matrix[m - i - 1][j]);
}
if (i == n - i - 1) { // only one col
continue;
}
// Traverse up
for (int j = m - i - 2; j > i; --j) {
list.add(matrix[j][i]);
}
}
return list;
}
}
