给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
示例 1:
输入: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] 输出: [1,2,3,6,9,8,7,4,5]
示例 2:
输入: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] 输出: [1,2,3,4,8,12,11,10,9,5,6,7]
从外往里一圈一圈遍历并存储矩阵元素即可。
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
def add(i1, j1, i2, j2):
if i1 == i2:
return [matrix[i1][j] for j in range(j1, j2 + 1)]
if j1 == j2:
return [matrix[i][j1] for i in range(i1, i2 + 1)]
return [matrix[i1][j] for j in range(j1, j2)] + [matrix[i][j2] for i in range(i1, i2)] + [matrix[i2][j] for j in range(j2, j1, -1)] + [matrix[i][j1] for i in range(i2, i1, -1)]
if not matrix or not matrix[0]:
return []
m, n = len(matrix), len(matrix[0])
i1, j1, i2, j2 = 0, 0, m - 1, n - 1
res = []
while i1 <= i2 and j1 <= j2:
res += add(i1, j1, i2, j2)
i1, j1, i2, j2 = i1 + 1, j1 + 1, i2 - 1, j2 - 1
return resclass Solution {
private List<Integer> res;
public List<Integer> spiralOrder(int[][] matrix) {
int m, n;
if (matrix == null || (m = matrix.length) == 0 || matrix[0] == null || (n = matrix[0].length) == 0)
return Collections.emptyList();
res = new ArrayList<>();
int i1 = 0, i2 = m - 1;
int j1 = 0, j2 = n - 1;
while (i1 <= i2 && j1 <= j2) {
add(matrix, i1++, j1++, i2--, j2--);
}
return res;
}
private void add(int[][] matrix, int i1, int j1, int i2, int j2) {
if (i1 == i2) {
for (int j = j1; j <= j2; ++j) {
res.add(matrix[i1][j]);
}
return;
}
if (j1 == j2) {
for (int i = i1; i <= i2; ++i) {
res.add(matrix[i][j1]);
}
return;
}
for (int j = j1; j < j2; ++j) {
res.add(matrix[i1][j]);
}
for (int i = i1; i < i2; ++i) {
res.add(matrix[i][j2]);
}
for (int j = j2; j > j1; --j) {
res.add(matrix[i2][j]);
}
for (int i = i2; i > i1; --i) {
res.add(matrix[i][j1]);
}
}
}