286.Walls-and-Gates

286. Walls and Gates

题目地址

https://leetcode.com/problems/walls-and-gates/

题目描述

You are given a m x n 2D grid initialized with these three possible values.

-1 - A wall or an obstacle.
0 - A gate.
INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.
Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

Example: 
Given the 2D grid:

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF
After running your function, the 2D grid should be:

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

代码

Approach #1 Multi End BFS

public static final int[] d = {0, 1, 0, -1, 0};

public void wallsAndGates(int[][] rooms) {
    if (rooms.length == 0) return;
    int m = rooms.length;
      int n = rooms[0].length;

    Deque<Integer> queue = new ArrayDeque<>();
    for (int i = 0; i < m; ++i)
        for (int j = 0; j < n; ++j)
            if (rooms[i][j] == 0) 
              queue.offer(i * n + j); // Put gates in the queue

    while (!queue.isEmpty()) {
        int x = queue.poll();
        int i = x / n, j = x % n;
        for (int k = 0; k < 4; ++k) {
            int p = i + d[k];
              int q = j + d[k + 1]; // empty room
            if (0 <= p && p < m && 0 <= q && q < n 
                && rooms[p][q] == Integer.MAX_VALUE) { // rooms[p][q] == Integer.MAX_VALUE

                rooms[p][q] = rooms[i][j] + 1;
                queue.offer(p * n + q);
            }
        }
    }
}

Approach #2 Naive BFS

public static final int[] d = {0, 1, 0, -1, 0};

public void wallsAndGates(int[][] rooms) {
    if (rooms.length == 0) return;
    for (int i = 0; i < rooms.length; ++i)
        for (int j = 0; j < rooms[0].length; ++j)
            if (rooms[i][j] == 0) 
              bfs(rooms, i, j);
}

private void bfs(int[][] rooms, int i, int j) {
    int m = rooms.length;
      int n = rooms[0].length;
    Deque<Integer> queue = new ArrayDeque<>();
    queue.offer(i * n + j); // Put gate in the queue
    while (!queue.isEmpty()) {
        int x = queue.poll();
        i = x / n; 
          j = x % n;
        for (int k = 0; k < 4; ++k) {
            int p = i + d[k];
              int q = j + d[k + 1];
            if (0 <= p && p < m && 0 <= q && q < n 
                && rooms[p][q] > rooms[i][j] + 1) { // rooms[p][q] > rooms[i][j] + 1

                rooms[p][q] = rooms[i][j] + 1;
                queue.offer(p * n + q);
            }
        }
    }
}

Approach #3 DFS

class Solution {
  int[] d = {0, 1, 0, -1, 0};

  public void wallsAndGates(int[][] rooms) {
    for (int i = 0; i < rooms.length; i++) {
      for (int j = 0; j < rooms[0].length; j++) {
        if (rooms[i][j] == 0) {
          dfs(rooms, i, j);
        }
      }
    }
  }

  public void dfs(int[][] rooms, int i, int j) {
    for (int k = 0; k < 5; k++) {
      int p = i + d[k];
      int q = j + d[k + 1];
      if (0 <= p && p < rooms.length 
          && 0 <= q && q < rooms[0].length 
          && rooms[p][q] > rooms[i][j] + 1) {

            rooms[p][q] = rooms[i][j] + 1;
            dfs(rooms, p, q);
        }
    }
  }

}

https://leetcode.com/problems/walls-and-gates/discuss/72748/Benchmarks-of-DFS-and-BFS