The Maze II

摘要：题目链接一般这种题，给一个起点，给一个终点，最后要求最短的路径，都是用来解。的图不是很大，也能。

题目链接：https://leetcode.com/contest/...
一般这种题，给一个起点，给一个终点，最后要求最短的路径，都是用bfs来解。

public class Solution {
    public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
        /* minheap: check the path with smaller length first
         * path[x][y]: record path from ball to (x, y)
         * visited[x][y]: length from ball to (x, y)
         */
        
        PriorityQueue heap = new PriorityQueue<>(new Comparator() {
            public int compare(Node a, Node b) {
                return a.len - b.len;
            }
        });
        heap.offer(new Node(ball[0], ball[1], 1));
        
        col = maze[0].length;
        
        // track result
        String result = "";
        int len = Integer.MAX_VALUE;
        
        // length and path so far
        visited = new int[maze.length][maze[0].length];
        visited[ball[0]][ball[1]] = 1;
        path = new String[maze.length][maze[0].length];
        path[ball[0]][ball[1]] = "";
        
        while(!heap.isEmpty()) {
            Node cur = heap.poll();
            visited[cur.x][cur.y] = cur.len;
            // find the hole, update result
            if(cur.x == hole[0] && cur.y == hole[1]) {
                if(len > cur.len || (len == cur.len && result.compareTo(path[cur.x][cur.y]) > 0)) {
                    len = cur.len;
                    result = path[cur.x][cur.y];
                } 
                continue;
            }
            // 4 directions
            for(int i = 0; i < 4; i++) {
                int nx = cur.x, ny = cur.y;
                int depth = cur.len;
                // find the next position (reach before the wall)
                while(nx + dx[i] >= 0 && nx + dx[i] < maze.length && ny + dy[i] >= 0 && ny + dy[i] < maze[0].length) {
                    // find the path so far from ball to [nx, ny]
                    String curPath = path[cur.x][cur.y] + (nx == cur.x && ny == cur.y ? "" : dir[i]);
                    // reach the hole
                    if(nx == hole[0] && ny == hole[1]) break;
                    // meet the wall
                    if(maze[nx + dx[i]][ny + dy[i]] == 1) break;
                    // loop update
                    depth++;
                    nx += dx[i];
                    ny += dy[i];
                }
                // pruning
                if(depth > len) continue;
                String curPath = path[cur.x][cur.y] + (nx == cur.x && ny == cur.y ? "" : dir[i]);
                if(visited[nx][ny] != 0 && visited[nx][ny] < depth) continue;
                if(visited[nx][ny] != 0 && visited[nx][ny] == depth && path[nx][ny].compareTo(curPath) <= 0) continue;
                // add new path
                visited[nx][ny]  = depth;
                path[nx][ny] = curPath;
                heap.offer(new Node(nx, ny, depth));
            }
        }
        return result.length() == 0 ? "impossible" : result;
    }
    
    int[] dx = new int[] {1, 0, 0, -1};
    int[] dy = new int[] {0, -1, 1, 0};
    char[] dir = new char[] {"d", "l", "r", "u"};
    
    int[][] visited;
    String[][] path;
    int col;
    
    class Node {
        int x, y;
        int len;
        Node(int x, int y, int len) {
            this.x = x;
            this.y = y;
            this.len = len;
        }
        @Override
        public int hashCode() {
            return this.x * col + this.y;
        }
        @Override
        public boolean equals(Object a) {
            if(a instaceof Node) {
                Node b = (Node) a;
                return b.x == this.x && b.y == this.y;
            }
            return false;
        }
    }
}

test的图不是很大，dfs也能ac。

public class Solution {
    public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
        /* use global variable: result and len 
         */
        visited = new boolean[maze.length][maze[0].length];
        dfs(maze, ball[0], ball[1], hole, 1, "");
        return result.length() == 0 ? "impossible" : result;
    }
    
    String result = "";
    int len = Integer.MAX_VALUE;
    boolean[][] visited;
    int[] dx = new int[] {1, 0, 0, -1};
    int[] dy = new int[] {0, -1, 1, 0};
    char[] dir = new char[] {"d", "l", "r", "u"};
    // dfs
    private void dfs(int[][] maze, int x, int y, int[] hole, int dep, String path) {
        if(dep > len) return;
        if(x == hole[0] && y == hole[1]) {
            if(len > dep) {
                len = dep;  result = path;
            }
            if(len == dep && path.compareTo(result) < 0) {
                len = dep;  result = path;
            } 
            return;
        }
        
        if(visited[x][y]) return;
        visited[x][y] = true;
        
        // 4 directions
        for(int i = 0; i < 4; i++) {
            int nx = x, ny = y;
            while(nx + dx[i] >= 0 && nx + dx[i] < maze.length && ny + dy[i] >= 0 && ny + dy[i] < maze[0].length) {
                // meet the wall
                if(maze[nx + dx[i]][ny + dy[i]] == 1) break;
                nx += dx[i];  ny += dy[i];
                if(visited[nx][ny]) break;
                if(nx == hole[0] && ny == hole[1]) break;
            }
            if(!visited[nx][ny]) dfs(maze, nx, ny, hole, dep + Math.abs(nx - x) + Math.abs(ny - y), path + dir[i]);
        }
        visited[x][y] = false;
    }
}