286 Walls and Gates
286. Walls and Gates
1. Question
You are given am x n2D 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 value2^31- 1 = 2147483647
to representINF
as you may assume that the distance to a gate is less than
2147483647
Fill each empty room with the distance to itsnearestgate. If it is impossible to reach a gate, it should be filled withINF
.
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
2. Implementation
(1) BFS
class Solution {
public void wallsAndGates(int[][] rooms) {
if (rooms == null || rooms.length == 0 || rooms[0].length == 0) {
return;
}
int m = rooms.length, n = rooms[0].length;
Queue<int[]> queue = new LinkedList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (rooms[i][j] == 0) {
queue.add(new int[] {i, j});
}
}
}
int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while (!queue.isEmpty()) {
int[] cell = queue.remove();
int curRow = cell[0], curCol = cell[1];
for (int[] direction : directions) {
int nextRow = curRow + direction[0];
int nextCol = curCol + direction[1];
if (isValid(rooms, nextRow, nextCol)) {
rooms[nextRow][nextCol] = rooms[curRow][curCol] + 1;
queue.add(new int[] {nextRow, nextCol});
}
}
}
}
public boolean isValid(int[][] matrix, int nextRow, int nextCol) {
return nextRow >= 0 && nextRow < matrix.length && nextCol >= 0 && nextCol < matrix[0].length && matrix[nextRow][nextCol] == Integer.MAX_VALUE;
}
}
3. Time & Space Complexity
BFS: 时间复杂度O(mn), 空间复杂度O(mn)
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