54. 螺旋矩阵【rust题解】

无名之逆2024-06-06 13:45

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

php 复制代码

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

rust 复制代码

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)
    }
}