力扣每日一题2021-11-03接雨水II

力扣每日一题2021-11-03接雨水II

接雨水II

• 407.接雨水II
• • 题目描述
  • 思路：广度优先搜索
  • • Java实现
    • Python实现

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407.接雨水II 题目描述

接雨水II

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思路：广度优先搜索

假定初始矩阵中的每个格子都装满了水，且高度均为maxHeight，其中maxHeight为矩阵中高度最高的格子。我们知道方块接水后的高度为water[i][j]，它的求解公式与方法一样，方块（i, j）的接水后的高度为： w a t e r [ i ] [ j ] = m a x ( h e i g h t M a p [ i ] , m i n ( w a t e r [ i − 1 ] [ j ] , w a t e r [ i + 1 ] [ j ] , w a t e r [ i ] [ j − 1 ] , w a t e r [ i ] [ j + 1 ] ) ) water[i][j] = max(heightMap[i], min(water[i-1][j], water[i+1][j], water[i][j-1], water[i][j+1])) water[i][j]=max(heightMap[i],min(water[i−1][j],water[i+1][j],water[i][j−1],water[i][j+1]))，而方块（i, j）实际接水的容量为water[i][j]-heightMap[i][j]。
假设每个方块（i, j）的接水后的高度均为water[i][j]=maxHeight，最外层的方块不能用于接水，所有的多余的水会从最外层的方块溢出，当前方块（i, j）的接水高度water[i][j]小于与之相邻的4个方块的接水高度时，则调整接水高度，将其相邻的四个方块的接水高度调整与（i, j）的高度保持一致，不断重复进行调整，直到所有的方块的接水高度不再有调整时即满足要求。

Java实现

class Solution {
    public int trapRainWater(int[][] heightMap) {
        int m = heightMap.length;
        int n = heightMap[0].length;
        int[] dirs = {-1, 0, 1, 0, -1};
        int maxHeight = 0;
        for (int i=0; i < m; ++i) {
            for (int j=0; j < n; ++j) {
                maxHeight = Math.max(maxHeight, heightMap[i][j]);
            }
        }
        int[][] water = new int[m][n];
        for (int i=0; i < m; ++i) {
            for (int j=0; j < n; ++j) {
                water[i][j] = maxHeight;
            }
        }
        Queue qu = new linkedList<>();
        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) {
                    if (water[i][j] > heightMap[i][j]) {
                        water[i][j] = heightMap[i][j];
                        qu.offer(new int[]{i ,j});
                    }
                }
            }
        }
        while (!qu.isEmpty()) {
            int[] cur = qu.poll();
            int x = cur[0];
            int y = cur[1];
            for (int i=0; i < 4; ++i) {
                int nx = x + dirs[i];
                int ny = y + dirs[i+1];
                if (nx < 0 || nx >= m || ny < 0 || ny >= n) {
                    continue;
                }
                if (water[x][y] < water[nx][ny] && water[nx][ny] > heightMap[nx][ny]) {
                    water[nx][ny] = Math.max(water[x][y], heightMap[nx][ny]);
                    qu.offer(new int[]{nx, ny});
                }
            }
        }
        int res = 0;
        for (int i = 0; i < m ; ++i) {
            for (int j = 0; j < n; ++j) {
                res += water[i][j] - heightMap[i][j];
            }
        }
        return res;
    }
}

Python实现

class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        m, n = len(heightMap), len(heightMap[0])
        maxHeight = max(max(row) for row in heightMap)
        water = [[maxHeight for i in range(n)] for j in range(m)]
        dirs = [-1, 0, 1, 0, -1]
        qu = []
        for i in range(m):
            for j in range(n):
                if i == 0 or i == m-1 or j == 0 or j == n-1:
                    if water[i][j] > heightMap[i][j]:
                        water[i][j] = heightMap[i][j]
                        qu.append([i, j])
        while len(qu) > 0:
            [x, y] = qu.pop(0)
            for i in range(4):
                nx, ny = x + dirs[i], y + dirs[i+1]
                if nx < 0 or nx >= m or ny < 0 or ny >= n:
                    continue
                if water[x][y] < water[nx][ny] and water[nx][ny] > heightMap[nx][ny]:
                    water[nx][ny] = max(water[x][y], heightMap[nx][ny])
                    qu.append([nx, ny])
        ans = 0
        for i in range(m):
            for j in range(n):
                ans = ans + water[i][j] - heightMap[i][j]
        return ans