317. Shortest Distance from All Buildings

题目：

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.

Example:

Input: [[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]]1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0Output: 7 Explanation: Given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2),
             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.

链接： http://leetcode.com/problems/maximum-product-of-word-lengths/

题解：

给一块空地，求在空地上新盖一座楼，到其他楼的距离最短。这里我们要用到BFS,就是从每个楼开始用BFS计算空地 “0”到这栋楼的距离，最后把每个空地到每栋楼的距离加起来，求一个最小值。这里我们还要算出楼的数目，仅当空地能连接所有楼的时候，我们才愿意在这块空地上造楼。我们也要维护一个visited矩阵来避免重复。

Time Complexity – O(4^mn)， Space Complexity – O(mn * k)， k为1的数目

public class Solution {
    private final int[][] directions = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};    public int shortestDistance(int[][] grid) {
        if(grid == null || grid.length == 0) {
            return Integer.MAX_VALUE;
        }
        int rowNum = grid.length;
        int colNum = grid[0].length;
        int[][] distance = new int[rowNum][colNum];
        int[][] canReachBuildings = new int[rowNum][colNum];
        int buildingNum = 0;         for(int i = 0; i < rowNum; i++) {
            for(int j = 0; j < colNum; j++) {
                if(grid[i][j] != 0) {
                    distance[i][j] = Integer.MAX_VALUE;
                }
                if(grid[i][j] == 1) {     // find out all buildings
                    buildingNum++;
                    updateDistance(grid, distance, canReachBuildings, i, j);
                }
            }
        }        int min = Integer.MAX_VALUE;
        for(int i = 0; i < rowNum; i++) {
            for(int j = 0; j < colNum; j++) {
                if(canReachBuildings[i][j] == buildingNum) {
                    min = Math.min(distance[i][j], min);
                }
            }
        }        return min == Integer.MAX_VALUE ? -1 : min;
    }    private void updateDistance(int[][] grid, int[][] distance, int[][] canReachBuildings, int row, int col) {
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{row, col});
        boolean[][] visited = new boolean[grid.length][grid[0].length];
        visited[row][col] = true;
        int dist = 0;
        int curLevel = 1;
        int nextLevel = 0;        while(!queue.isEmpty()) {
            int[] position = queue.poll();
            distance[position[0]][position[1]] += dist;
            curLevel--;
            for(int[] direction : directions) {
                int x = position[0] + direction[0];
                int y = position[1] + direction[1];
                if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] != 0) {
                    continue;
                }
                if(!visited[x][y]) {
                    queue.offer(new int[]{x, y});
                    nextLevel++;
                    visited[x][y] = true;
                    canReachBuildings[x][y]++;
                }
            }
            if(curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                dist++;
            }
        }
    }
}

Reference:

https://leetcode.com/discuss/74453/36-ms-c-solution

https://leetcode.com/discuss/74422/clean-solution-easy-understanding-with-simple-explanation

https://leetcode.com/discuss/74999/java-solution-with-explanation-and-time-complexity-analysis

https://leetcode.com/discuss/74380/my-bfs-java-solution