# 286.Walls-and-Gates

## ้ข็ฎๅฐๅ

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

## ้ข็ฎๆ่ฟฐ

``````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``````

## ไปฃ็ 

### 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

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