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[LeetCode]Shortest Distance from All Buildings

2015-12-14 14:02 676 查看
对每个房子bfs,找出每个房子到所有空地的距离。用一个record二维数组来记录每个空地上所有房子到该空地的距离和。同时用另一个二维数组记录每个空地能被几个房子访问到,来保证所有房子都可以到达的了该空地。

public class Solution {
public int shortestDistance(int[][] grid) {
int row = grid.length;
if (row == 0) {
return -1;
}
int col = grid[0].length;
int[][] record1 = new int[row][col]; // visited num
int[][] record2 = new int[row][col]; // distance
int num1 = 0;
for (int r = 0; r < row; r++) {
for (int c = 0; c < col; c++) {
if (grid[r][c] == 1) {
num1 ++;
boolean[][] visited = new boolean[row][col];
Queue<int[]> queue = new LinkedList<int[]>();
queue.offer(new int[]{r, c});
int dist = 0;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] node = queue.poll();
int x = node[0];
int y = node[1];
record2[x][y] += dist;
record1[x][y] ++;
if (x > 0 && grid[x - 1][y] == 0 && !visited[x - 1][y]) {
queue.offer(new int[]{x - 1, y});
visited[x - 1][y] = true;
}
if (x + 1 < row && grid[x + 1][y] == 0 && !visited[x + 1][y]) {
queue.offer(new int[]{x + 1, y});
visited[x + 1][y] = true;
}
if (y > 0 && grid[x][y - 1] == 0 && !visited[x][y - 1]) {
queue.offer(new int[]{x, y - 1});
visited[x][y - 1] = true;
}
if (y + 1 < col && grid[x][y + 1] == 0 && !visited[x][y + 1]) {
queue.offer(new int[]{x, y + 1});
visited[x][y + 1] = true;
}
}
dist ++;
}
}
}
}
int result = Integer.MAX_VALUE;
for (int r = 0; r < row; r++) {
for (int c = 0; c < col; c++) {
if (grid[r][c] == 0 && record1[r][c] == num1 && record2[r][c] < result) {
result = record2[r][c];
}
}
}
return result == Integer.MAX_VALUE ? -1 : result;
}
}
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