[LeetCode]Shortest Distance from All Buildings

Shortest Distance from All Buildings

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

• Each 0 marks an empty land which you can pass by freely.

• Each 1 marks a building which you cannot pass through.

• Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

分析

這道理相似於Walls ang Gates, 解決方法也應該是從building出發進行BFS，不過這裏不一樣的是這裏須要返回最小距離和，因此咱們應該一個一個的對building的點BFS，用一個二維矩陣存每一個點到全部building的距離和，每次BFS，都更新相應的距離和。最後遍歷那個距離和矩陣，找出最小值便可。

須要注意的是，這道題還有個條件就是empty room 必須 reach all buildings，因此咱們能夠用另一個矩陣存對應empty room到building的個數，若是最終個數不等於總的building數，對應點存的距離和無效。

複雜度

time: O(kMN), space: O(MN), k表示building數量

代碼

public class Solution {
    public int shortestDistance(int[][] grid) {
        int rows = grid.length;
        if (rows == 0) {
            return -1;
        }
        int cols = grid[0].length;
 
        // 記錄到各個building距離和
        int[][] dist = new int[rows][cols];
        
        // 記錄到能到達的building的數量
        int[][] nums = new int[rows][cols];            
        int buildingNum = 0;
        
        // 從每一個building開始BFS
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1) {
                    buildingNum++;
                    bfs(grid, i, j, dist, nums);
                }
            }
        }
        
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 0 && dist[i][j] != 0 && nums[i][j] == buildingNum)
                    min = Math.min(min, dist[i][j]);
            }
        }
        if (min < Integer.MAX_VALUE)
            return min;
        return -1;
    }
    
    public void bfs(int[][] grid, int row, int col, int[][] dist, int[][] nums) {
        int rows = grid.length;
        int cols = grid[0].length;
        
        Queue<int[]> q = new LinkedList<>();
        q.add(new int[]{row, col});
        int[][] dirs = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        
        // 記錄訪問過的點
        boolean[][] visited = new boolean[rows][cols];
        int level = 0;
        while (!q.isEmpty()) {
            level++;
            int size = q.size();
            for (int i = 0; i < size; i++) {
                int[] coords = q.remove();
                for (int k = 0; k < dirs.length; k++) {
                    int x = coords[0] + dirs[k][0];
                    int y = coords[1] + dirs[k][1];
                    if (x >= 0 && x < rows && y >= 0 && y < cols && !visited[x][y] && grid[x][y] == 0) {
                        visited[x][y] = true;
                        dist[x][y] += level;
                        nums[x][y]++;
                        q.add(new int[]{x, y});
                    }
                }
            }
        }
    }
}