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。
输入: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] 输出: [1,2,3,6,9,8,7,4,5]
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]
impl Solution {
pub fn spiral_order(matrix: Vec<Vec<i32>>) -> Vec<i32> {
if matrix.len() == 0 || matrix[0].len() == 0 {
return Vec::new();
}
let mut row = 0;
let mut col = 0;
let mut u = 0;
let mut d = matrix.len() as i32 - 1;
let mut l = 0;
let mut r = matrix[0].len() as i32 - 1;
let mut dir = (0, 1);
let mut ret = Vec::new();
for _ in 0..(matrix.len() * matrix[0].len()) {
ret.push(matrix[row as usize][col as usize]);
match dir {
(0, 1) if col == r => { dir = (1, 0); u += 1; },
(1, 0) if row == d => { dir = (0, -1); r -= 1; },
(0, -1) if col == l => { dir = (-1, 0); d -= 1; },
(-1, 0) if row == u => { dir = (0, 1); l += 1; },
_ => (),
}
row += dir.0;
col += dir.1;
}
ret
}
}