The Maze II

https://leetcode.com/problems/the-maze-ii/description/

There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

Given the ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1.

The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.

Example 1

Input 1: a maze represented by a 2D array

0 0 1 0 0

0 0 0 0 0

0 0 0 1 0

1 1 0 1 1

0 0 0 0 0

Input 2: start coordinate (rowStart, colStart) = (0, 4)

Input 3: destination coordinate (rowDest, colDest) = (4, 4)

Output: 12

Explanation: One shortest way is : left -> down -> left -> down -> right -> down -> right.

         The total distance is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.

Input 1: a maze represented by a 2D array

0 0 1 0 0

0 0 0 0 0

0 0 0 1 0

1 1 0 1 1

0 0 0 0 0

Input 2: start coordinate (rowStart, colStart) = (0, 4)

Input 3: destination coordinate (rowDest, colDest) = (3, 2)

Output: -1

Explanation: There is no way for the ball to stop at the destination.

Note:

There is only one ball and one destination in the maze.

Both the ball and the destination exist on an empty space, and they will not be at the same position initially.

The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls.

The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.

Thoughts

刚开始想的解法有点复杂, 想用四个hashmap把不同方向到每个点的距离记下来. 实际上不需要每个访问过的点都记录, 只需要记录每次rolling后的终点离起点的距离即可, 相当于记录了所有中间节点的距离.

Code

class Solution {
public:
    int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
        int m = maze.size();
        if (m == 0) return 0;
        int n = maze[0].size();
        vector<vector<int>> dirs({{0, 1}, {0, -1}, {-1, 0}, {1, 0}});
        queue<vector<int>> q;
        q.push(start);
        vector<vector<int>> visited(m, vector<int>(n, INT_MAX));
        visited[start[0]][start[1]] = 0;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                const auto p = q.front();
                q.pop();
                for (const auto& dir : dirs) {
                    auto x = p[0] + dir[0];
                    auto y = p[1] + dir[1];
                    auto d = visited[p[0]][p[1]] + 1;
                    while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == 0) {
                        x += dir[0];
                        y += dir[1];
                        ++d;
                    }
                    x -= dir[0];
                    y -= dir[1];
                    --d;
                    if (d < visited[x][y]) {
                        q.push({x, y});
                        visited[x][y] = d;
                    }
                }
            }
        }
        const auto res = visited[destination[0]][destination[1]]; 
        return res == INT_MAX ? -1 : res;
    }
};

class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] destination) {
        int m = maze.length, n = m == 0 ? 0 : maze[0].length;
        int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int[][] dists = new int[m][n];

        for (int i = 0; i < m * n; i++) {
            dists[i / n][i % n] = Integer.MAX_VALUE;
        }

        Queue<int[]> queue = new LinkedList<>();
        dists[start[0]][start[1]] = 0;
        queue.offer(start);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] pos = queue.poll();
                int dist = dists[pos[0]][pos[1]];
                for (int[] dir : dirs) {
                    int ndist = dist + 1;
                    int x = pos[0] + dir[0];
                    int y = pos[1] + dir[1];
                    while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == 0) {
                        x += dir[0];
                        y += dir[1];
                        ndist++;
                    }
                    x -= dir[0];
                    y -= dir[1];
                    ndist--;
                    if (dists[x][y] > ndist) {
                        dists[x][y] = ndist;
                        queue.offer(new int[]{x, y});
                    }
                }
            }
        }

        return dists[destination[0]][destination[1]] == Integer.MAX_VALUE ? -1 : dists[destination[0]][destination[1]];
    }
}

Analysis

时间复杂度O(MN), 空间O(MN).