java面试题(1)

原题：

// 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.

答案:

public class Solution {    public int shortestDistance(int[][] grid) {        if(grid == null || grid.length == 0 || grid[0].length == 0) return -1;        final int[] shift = {0, 1, 0, -1, 0};        int rows = grid.length;        int columns = grid[0].length;        int[][] distance = new int[rows][columns];        int[][] reach = new int[rows][columns];        int numberOfBuildings = 0;        for(int i = 0; i < rows; i++) {            for(int j = 0; j < columns; j++) {                if(grid[i][j] == 1) {                    numberOfBuildings++;                    Queue<int[]> queue = new LinkedList<int[]>();                    queue.offer(new int[] {i, j});                    boolean[][] visited = new boolean[rows][columns];                    int relativeDistance = 1;                    while(!queue.isEmpty()) {                        int qSize = queue.size();                        for(int q = 0; q < qSize; q++) {                            int[] current = queue.poll();                            for(int k = 0; k < 4; k++) {                                int nextRow = current[0] + shift[k];                                int nextColumn = current[1] + shift[k + 1];                                if(nextRow >= 0 && nextRow < rows && nextColumn >= 0 && nextColumn < columns && grid[nextRow][nextColumn] == 0 && !visited[nextRow][nextColumn]) {                                    distance[nextRow][nextColumn] += relativeDistance;                                    reach[nextRow][nextColumn]++;                                    visited[nextRow][nextColumn] = true;                                    queue.offer(new int[] {nextRow, nextColumn});                                }                               }                        }                        relativeDistance++;                    }                }            }        }        int shortest = Integer.MAX_VALUE;        for(int i = 0; i < rows; i++) {            for(int j = 0; j < columns; j++) {                if(grid[i][j] == 0 && reach[i][j] == numberOfBuildings) {                    shortest = Math.min(shortest, distance[i][j]);                }            }        }        return shortest == Integer.MAX_VALUE ? -1 : shortest;    }}