LC.P1765[地图中的最高点]

方法一:多源BFS

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
class Solution {
static int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

public int[][] highestPeak(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] ans = new int[m][n];
Deque<int[]> queue = new ArrayDeque<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) queue.offer(new int[]{i, j});
ans[i][j] = grid[i][j] == 1 ? 0 : -1;
}
}
while (!queue.isEmpty()) {
int[] cur = queue.poll();
int x = cur[0], y = cur[1];
int height = ans[x][y];
for (int[] dir : dirs) {
int nx = x + dir[0], ny = y + dir[1];
if (nx < 0 || nx >= m || ny < 0 || ny >= n) continue;
if (ans[nx][ny] != -1) continue;
ans[nx][ny] = height + 1;
queue.offer(new int[]{nx, ny});
}
}
return ans;
}
}
  • 时间复杂度:$O(m \times n)$
  • 空间复杂度:$O(m \times n)$