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# 286.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.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.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.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.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.length
&& rooms[p][q] > rooms[i][j] + 1) {
rooms[p][q] = rooms[i][j] + 1;
dfs(rooms, p, q);
}
}
}
}