LeetCode每日一题(1706. Where Will the Ball Fall)

You have a 2-D grid of size m x n representing a box, and you have n balls. The box is open on the top and bottom sides.

Each cell in the box has a diagonal board spanning two corners of the cell that can redirect a ball to the right or to the left.

A board that redirects the ball to the right spans the top-left corner to the bottom-right corner and is represented in the grid as 1.
A board that redirects the ball to the left spans the top-right corner to the bottom-left corner and is represented in the grid as -1.
We drop one ball at the top of each column of the box. Each ball can get stuck in the box or fall out of the bottom. A ball gets stuck if it hits a “V” shaped pattern between two boards or if a board redirects the ball into either wall of the box.

Return an array answer of size n where answer[i] is the column that the ball falls out of at the bottom after dropping the ball from the ith column at the top, or -1 if the ball gets stuck in the box.

Example 1:

Input: grid = [[1,1,1,-1,-1],[1,1,1,-1,-1],[-1,-1,-1,1,1],[1,1,1,1,-1],[-1,-1,-1,-1,-1]]
Output: [1,-1,-1,-1,-1]

Explanation: This example is shown in the photo.
Ball b0 is dropped at column 0 and falls out of the box at column 1.
Ball b1 is dropped at column 1 and will get stuck in the box between column 2 and 3 and row 1.
Ball b2 is dropped at column 2 and will get stuck on the box between column 2 and 3 and row 0.
Ball b3 is dropped at column 3 and will get stuck on the box between column 2 and 3 and row 0.
Ball b4 is dropped at column 4 and will get stuck on the box between column 2 and 3 and row 1.

Example 2:

Input: grid = [[-1]]
Output: [-1]

Explanation: The ball gets stuck against the left wall.

Example 3:

Input: grid = [[1,1,1,1,1,1],[-1,-1,-1,-1,-1,-1],[1,1,1,1,1,1],[-1,-1,-1,-1,-1,-1]]
Output: [0,1,2,3,4,-1]

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• grid[i][j] is 1 or -1.

---

别想复杂了， 这题就没什么难度。假设 grid[r][c] == 1(右下), 只要 grid[r][c+1]1 就可以进入下一行， 如果 grid[r][c] == -1(左下), 只要 grid[r][c-1]-1 就可以进入下一行。 最后一行的规则跟这一样， 只不过不会进入下一行而是直接掉出。注意考虑两边是墙的情况。

---

impl Solution {
    fn check(grid: &Vec<Vec<i32>>, row: usize, col: usize) -> i32 {
        let curr = grid[row][col];
        if row == grid.len() - 1 {
            if curr == 1 {
                if col == grid[0].len() - 1 {
                    return -1;
                }
                let right = grid[row][col + 1];
                if right == -1 {
                    return -1;
                }
                return col as i32 + 1;
            }
            if col == 0 {
                return -1;
            }
            let left = grid[row][col - 1];
            if left == 1 {
                return -1;
            }
            return col as i32 - 1;
        }
        if curr == -1 {
            if col == 0 {
                return -1;
            }
            let left = grid[row][col - 1];
            if left == 1 {
                return -1;
            }
            return Solution::check(grid, row + 1, col - 1);
        }
        if col == grid[0].len() - 1 {
            return -1;
        }
        let right = grid[row][col + 1];
        if right == -1 {
            return -1;
        }
        return Solution::check(grid, row + 1, col + 1);
    }
    pub fn find_ball(grid: Vec<Vec<i32>>) -> Vec<i32> {
        let mut ans = vec![-1; grid[0].len()];
        for i in 0..ans.len() {
            let idx = Solution::check(&grid, 0, i);
            if idx >= 0 {
                ans[i] = idx;
            }
        }
        ans
    }
}
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