286. Walls and Gates

题目：

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

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. 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 than2147483647.

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.

For 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

链接： http://leetcode.com/problems/walls-and-gates/

题解：

求矩阵中room到gate的最短距离。这里我们可以用BFS， 先把所有gate放入queue中，再根据gate计算上下左右，假如有room,再把这个room加入到queue中。

Time Complexity - O(4ⁿ)， Space Complexity - O(4ⁿ)

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        if(rooms == null || rooms.length == 0) {
            return;
        }
 
        Queue<int[]> queue = new LinkedList<>();
        for(int i = 0; i < rooms.length; i++) {
            for(int j = 0; j < rooms[0].length; j++) {
                if(rooms[i][j] == 0) {
                    queue.add(new int[]{i, j});
                }
            }
        }
        while(!queue.isEmpty()) {
            int[] gate = queue.poll();
            int row = gate[0], col = gate[1];
            if(row > 0 && rooms[row - 1][col] == Integer.MAX_VALUE) {
                rooms[row - 1][col] = rooms[row][col] + 1;
                queue.add(new int[]{row - 1, col});
            }
            if(row < rooms.length - 1 && rooms[row + 1][col] == Integer.MAX_VALUE) {
                rooms[row + 1][col] = rooms[row][col] + 1;
                queue.add(new int[]{row + 1, col});
            }
            if(col > 0 && rooms[row][col - 1] == Integer.MAX_VALUE) {
                rooms[row][col - 1] = rooms[row][col] + 1;
                queue.add(new int[]{row, col - 1});
            }
            if(col < rooms[0].length - 1 && rooms[row][col + 1] == Integer.MAX_VALUE) {
                rooms[row][col + 1] = rooms[row][col] + 1;
                queue.add(new int[]{row, col + 1});
            }
        }
    }
}

二刷:

用了另外一种写法。但两种速度都不是很快。

Java:

public class Solution {
    private int[][] directions = new int[][] {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
 
    public void wallsAndGates(int[][] rooms) {
        if (rooms == null || rooms.length == 0) {
            return;
        }
        Queue<int[]> queue = new LinkedList<>();
        int dist = 1, curLevel = 0, nextLevel = 0;
        for (int i = 0; i < rooms.length; i++) {
            for (int j = 0; j < rooms[0].length; j++) {
                if (rooms[i][j] == 0) {
                    queue.offer(new int[] {i, j});
                    curLevel++;
                }
            }
        }
 
        while (!queue.isEmpty()) {
            int[] gate = queue.poll();
            curLevel--;
            for (int[] direction : directions) {
                int row = direction[0] + gate[0];
                int col = direction[1] + gate[1];
                if (row < 0
                    || row > rooms.length - 1
                    || col < 0
                    || col > rooms[0].length - 1
                    || rooms[row][col] <= 0
                    || rooms[row][col] <= dist) {
                    continue;
                }
                rooms[row][col] = dist;
                queue.offer(new int[] {row, col});
                nextLevel++;
            }
            if (curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                dist++;
            }
        }
    }
}

又是参考yavinci大神的DFS，代码简洁速度也快

public class Solution {
    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, 0);
                }
            }
        }
    }
 
    public void dfs(int[][] rooms, int i, int j, int d) {
        if (i < 0 || i >= rooms.length || j < 0 || j >= rooms[0].length || rooms[i][j] < d) {
            return;
        }
        rooms[i][j] = d;
        dfs(rooms, i - 1, j, d + 1);
        dfs(rooms, i, j - 1, d + 1);
        dfs(rooms, i + 1, j, d + 1);
        dfs(rooms, i, j + 1, d + 1);
    }
}

Reference:

https://leetcode.com/discuss/68456/java-easiest-dfs-quicker-than-bfs

https://www.cs.ubc.ca/~kevinlb/teaching/cs322%20-%202008-9/Lectures/Search3.pdf

https://en.wikipedia.org/wiki/Breadth-first_search#Time_and_space_complexity

https://leetcode.com/discuss/60552/beautiful-java-solution-10-lines

https://leetcode.com/discuss/60418/c-bfs-clean-solution-with-simple-explanations

https://leetcode.com/discuss/60170/6-lines-o-mn-python-bfs

https://leetcode.com/discuss/73686/concise-java-solution-bfs-7ms

https://leetcode.com/discuss/68456/java-dfs-solution-much-quicker-and-simpler-than-bfs