LC.P407[接雨水II]

方法一:优先队列

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class Solution {
private static final int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

public int trapRainWater(int[][] heights) {
int m = heights.length, n = heights[0].length;
boolean[][] visited = new boolean[m][n];
// 存放三元组 [x,y,h]
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[2] - b[2]);
// 先把最外一圈放进去
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
pq.offer(new int[]{i, j, heights[i][j]});
visited[i][j] = true;
}
}
}
int ans = 0;
while (!pq.isEmpty()) {
int[] cur = pq.poll();
for (int[] dir : dirs) {
int nx = cur[0] + dir[0], ny = cur[1] + dir[1];
if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny]) {
if (cur[2] > heights[nx][ny]) {
ans += cur[2] - heights[nx][ny];
}
pq.offer(new int[]{nx, ny, Math.max(heights[nx][ny], cur[2])});
visited[nx][ny] = true;
}
}
}
return ans;
}
}
  • 时间复杂度:$O(m^2n^2)$
  • 空间复杂度:$O(mn)$