张家界网站建设多少钱平台推广是什么工作

题目

给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

示例

示例 1

输入：matrix = [[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：matrix = [[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]

提示

m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

思路

深搜，方向旋转。需要特别处理在最左边往上搜索的情况。

AC代码

impl Solution {pub fn dfs(v: &Vec<Vec<i32>>, vis: &mut Vec<Vec<bool>>, x: i32, y: i32) -> Vec<i32> {let x_len = v.len() as i32;let y_len = v[0].len() as i32;let mut res: Vec<i32> = Vec::new();if x < 0 || x >= x_len || y < 0 || y >= y_len || vis[x as usize][y as usize] {return res;}vis[x as usize][y as usize] = true;let mut res1: Vec<i32> = Vec::new();let mut res2: Vec<i32> = Vec::new();let mut res3: Vec<i32> = Vec::new();let mut res4: Vec<i32> = Vec::new();res1 = match x >= 1 && (y == 0 || (y >= 1 && vis[x as usize][y as usize - 1])) && !vis[x as usize - 1][y as usize]{true => Solution::dfs(v, vis, x - 1, y),_ => Solution::dfs(v, vis, x, y + 1)};res2 = Solution::dfs(v, vis, x + 1, y);res3 = Solution::dfs(v, vis, x, y - 1);res4 = Solution::dfs(v, vis, x - 1, y);res.push(v[x as usize][y as usize]);res.extend(res1);res.extend(res2);res.extend(res3);res.extend(res4);res}pub fn spiral_order(v: Vec<Vec<i32>>) -> Vec<i32> {let x_len = v.len();let y_len = v[0].len();let mut vis: Vec<Vec<bool>> = vec![vec![false; y_len]; x_len];Solution::dfs(&v, &mut vis, 0 , 0)}
}