Q54 螺旋矩阵
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]
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> list = new ArrayList<>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return list;
int lr = 0;
int lc = 0;
int rr = matrix.length - 1;
int rc = matrix[0].length - 1;
while (lr <= rr && lc <= rc)
print(matrix, lr++, lc++, rr--, rc--, list);
return list;
}
private void print(int[][] matrix, int lr, int lc, int rr, int rc, List<Integer> list) {
if (lr == rr) {
for (int i = lc; i <= rc; i++)
list.add(matrix[lr][i]);
} else if (lc == rc) {
for (int i = lr; i <= rr; i++)
list.add(matrix[i][lc]);
} else {
for (int i = lc; i < rc; i++)
list.add(matrix[lr][i]);
for (int i = lr; i < rr; i++)
list.add(matrix[i][rc]);
for (int i = rc; i > lc; i--)
list.add(matrix[rr][i]);
for (int i = rr; i > lr; i--)
list.add(matrix[i][lc]);
}
}
}