Leetcode_1293_Shortest Path in a Grid with Obstacles Elimination

[lihe]

Shortest Path in a Grid with Obstacles Elimination

Given a m * n grid, where each cell is either 0 (empty) or 1 (obstacle). In one step, you can move up, down, left or right from and to an empty cell.

Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m-1, n-1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.

Example 1:

Input: 
grid = 
[[0,0,0],
 [1,1,0],
 [0,0,0],
 [0,1,1],
 [0,0,0]], 
k = 1
Output: 6
Explanation: 
The shortest path without eliminating any obstacle is 10. 
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

Example 2:

Input: 
grid = 
[[0,1,1],
 [1,1,1],
 [1,0,0]], 
k = 1
Output: -1
Explanation: 
We need to eliminate at least two obstacles to find such a walk.

Constraints:

• grid.length == m
• grid[0].length == n
• 1 <= m, n <= 40
• 1 <= k <= m*n
• grid[i][j] == 0 **or** 1
• grid[0][0] == grid[m-1][n-1] == 0

算法思路：利用BFS解题，将二维矩阵扩充为三维，z代表当前消除的障碍物数量；visited[x][y][z]，代表消除z个障碍是否到达[x, y]处。

class Solution {
    private int[][] dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};


    public int shortestPath(int[][] grid, int k) {
        int m = grid.length;
        int n = grid[0].length;

        boolean[][][] visited = new boolean[m][n][k + 1];
        Queue<Point> queue = new LinkedList<>();

        queue.add(new Point(0, 0, 0));

        visited[0][0][0] = true;

        int distance = 0;

        while (!queue.isEmpty()){
            distance++;

            int size = queue.size();

            for(int i = 0; i < size; i++){
                Point point = queue.poll();

                int x = point.x;
                int y = point.y;
                int z = point.z;

                if(x == m - 1 && y == n - 1){  // 达到目的地
                    return distance - 1;
                }

                for(int j = 0; j < 4; j++){  // 四个方向
                    int new_x = x + dir[j][0];
                    int new_y = y + dir[j][1];
                    int new_z = z;

                    if(new_x < 0 || new_x >= m || new_y < 0 || new_y >= n){  // 超出矩阵范围
                        continue;
                    }

                    if(grid[new_x][new_y] == 1){
                        if ( z < k){  // 可以继续消除障碍物
                            new_z = z + 1;
                        }
                        else{
                            continue;  // 已经到达上限
                        }
                    }

                    if(!visited[new_x][new_y][new_z]){  // 判断是否访问过
                        queue.add(new Point(new_x, new_y, new_z));

                        visited[new_x][new_y][new_z] = true;
                    }
                }
            }
        }
        return -1;
    }

    class Point{
        int x;
        int y;
        int z;

        public Point(int x, int y, int z){
            this.x = x;
            this.y = y;
            this.z = z;
        }
    }
}