Question

417. Pacific Atlantic Water Flow

中文文档

Description

There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches the island's right and bottom edges.

The island is partitioned into a grid of square cells. You are given an m x n integer matrix heights where heights[r][c] represents the height above sea level of the cell at coordinate (r, c).

The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is less than or equal to the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.

Return _a 2D list of grid coordinates _result_ where _result[i] = [ri, ci]_ denotes that rain water can flow from cell _(ri, ci)_ to both the Pacific and Atlantic oceans_.

 

Example 1:

Input: heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
Output: [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]
Explanation: The following cells can flow to the Pacific and Atlantic oceans, as shown below:
[0,4]: [0,4] -> Pacific Ocean 
     [0,4] -> Atlantic Ocean
[1,3]: [1,3] -> [0,3] -> Pacific Ocean 
     [1,3] -> [1,4] -> Atlantic Ocean
[1,4]: [1,4] -> [1,3] -> [0,3] -> Pacific Ocean 
     [1,4] -> Atlantic Ocean
[2,2]: [2,2] -> [1,2] -> [0,2] -> Pacific Ocean 
     [2,2] -> [2,3] -> [2,4] -> Atlantic Ocean
[3,0]: [3,0] -> Pacific Ocean 
     [3,0] -> [4,0] -> Atlantic Ocean
[3,1]: [3,1] -> [3,0] -> Pacific Ocean 
     [3,1] -> [4,1] -> Atlantic Ocean
[4,0]: [4,0] -> Pacific Ocean 
     [4,0] -> Atlantic Ocean
Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.

Example 2:

Input: heights = [[1]]
Output: [[0,0]]
Explanation: The water can flow from the only cell to the Pacific and Atlantic oceans.

 

Constraints:

• m == heights.length
• n == heights[r].length
• 1 <= m, n <= 200
• 0 <= heights[r][c] <= 105

Solutions

Solution 1

class Solution:
  def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
    def bfs(q, vis):
      while q:
        for _ in range(len(q)):
          i, j = q.popleft()
          for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
            x, y = i + a, j + b
            if (
              0 <= x < m
              and 0 <= y < n
              and (x, y) not in vis
              and heights[x][y] >= heights[i][j]
            ):
              vis.add((x, y))
              q.append((x, y))

    m, n = len(heights), len(heights[0])
    vis1, vis2 = set(), set()
    q1 = deque()
    q2 = deque()
    for i in range(m):
      for j in range(n):
        if i == 0 or j == 0:
          vis1.add((i, j))
          q1.append((i, j))
        if i == m - 1 or j == n - 1:
          vis2.add((i, j))
          q2.append((i, j))
    bfs(q1, vis1)
    bfs(q2, vis2)
    return [
      (i, j)
      for i in range(m)
      for j in range(n)
      if (i, j) in vis1 and (i, j) in vis2
    ]

class Solution {
  private int[][] heights;
  private int m;
  private int n;

  public List<List<Integer>> pacificAtlantic(int[][] heights) {
    m = heights.length;
    n = heights[0].length;
    this.heights = heights;
    Deque<int[]> q1 = new LinkedList<>();
    Deque<int[]> q2 = new LinkedList<>();
    Set<Integer> vis1 = new HashSet<>();
    Set<Integer> vis2 = new HashSet<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == 0 || j == 0) {
          vis1.add(i * n + j);
          q1.offer(new int[] {i, j});
        }
        if (i == m - 1 || j == n - 1) {
          vis2.add(i * n + j);
          q2.offer(new int[] {i, j});
        }
      }
    }
    bfs(q1, vis1);
    bfs(q2, vis2);
    List<List<Integer>> ans = new ArrayList<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int x = i * n + j;
        if (vis1.contains(x) && vis2.contains(x)) {
          ans.add(Arrays.asList(i, j));
        }
      }
    }
    return ans;
  }

  private void bfs(Deque<int[]> q, Set<Integer> vis) {
    int[] dirs = {-1, 0, 1, 0, -1};
    while (!q.isEmpty()) {
      for (int k = q.size(); k > 0; --k) {
        int[] p = q.poll();
        for (int i = 0; i < 4; ++i) {
          int x = p[0] + dirs[i];
          int y = p[1] + dirs[i + 1];
          if (x >= 0 && x < m && y >= 0 && y < n && !vis.contains(x * n + y)
            && heights[x][y] >= heights[p[0]][p[1]]) {
            vis.add(x * n + y);
            q.offer(new int[] {x, y});
          }
        }
      }
    }
  }
}

typedef pair<int, int> pii;

class Solution {
public:
  vector<vector<int>> heights;
  int m;
  int n;

  vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
    m = heights.size();
    n = heights[0].size();
    this->heights = heights;
    queue<pii> q1;
    queue<pii> q2;
    unordered_set<int> vis1;
    unordered_set<int> vis2;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == 0 || j == 0) {
          vis1.insert(i * n + j);
          q1.emplace(i, j);
        }
        if (i == m - 1 || j == n - 1) {
          vis2.insert(i * n + j);
          q2.emplace(i, j);
        }
      }
    }
    bfs(q1, vis1);
    bfs(q2, vis2);
    vector<vector<int>> ans;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int x = i * n + j;
        if (vis1.count(x) && vis2.count(x)) {
          ans.push_back({i, j});
        }
      }
    }
    return ans;
  }

  void bfs(queue<pii>& q, unordered_set<int>& vis) {
    vector<int> dirs = {-1, 0, 1, 0, -1};
    while (!q.empty()) {
      for (int k = q.size(); k > 0; --k) {
        auto p = q.front();
        q.pop();
        for (int i = 0; i < 4; ++i) {
          int x = p.first + dirs[i];
          int y = p.second + dirs[i + 1];
          if (x >= 0 && x < m && y >= 0 && y < n && !vis.count(x * n + y) && heights[x][y] >= heights[p.first][p.second]) {
            vis.insert(x * n + y);
            q.emplace(x, y);
          }
        }
      }
    }
  }
};

func pacificAtlantic(heights [][]int) [][]int {
  m, n := len(heights), len(heights[0])
  vis1 := make(map[int]bool)
  vis2 := make(map[int]bool)
  var q1 [][]int
  var q2 [][]int
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      if i == 0 || j == 0 {
        vis1[i*n+j] = true
        q1 = append(q1, []int{i, j})
      }
      if i == m-1 || j == n-1 {
        vis2[i*n+j] = true
        q2 = append(q2, []int{i, j})
      }
    }
  }
  dirs := []int{-1, 0, 1, 0, -1}
  bfs := func(q [][]int, vis map[int]bool) {
    for len(q) > 0 {
      for k := len(q); k > 0; k-- {
        p := q[0]
        q = q[1:]
        for i := 0; i < 4; i++ {
          x, y := p[0]+dirs[i], p[1]+dirs[i+1]
          if x >= 0 && x < m && y >= 0 && y < n && !vis[x*n+y] && heights[x][y] >= heights[p[0]][p[1]] {
            vis[x*n+y] = true
            q = append(q, []int{x, y})
          }
        }
      }
    }
  }
  bfs(q1, vis1)
  bfs(q2, vis2)
  var ans [][]int
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      x := i*n + j
      if vis1[x] && vis2[x] {
        ans = append(ans, []int{i, j})
      }
    }
  }
  return ans
}

function pacificAtlantic(heights: number[][]): number[][] {
  const m = heights.length;
  const n = heights[0].length;
  const dirs = [
    [1, 0],
    [0, 1],
    [-1, 0],
    [0, -1],
  ];
  const gird = new Array(m).fill(0).map(() => new Array(n).fill(0));
  const isVis = new Array(m).fill(0).map(() => new Array(n).fill(false));

  const dfs = (i: number, j: number) => {
    if (isVis[i][j]) {
      return;
    }
    gird[i][j]++;
    isVis[i][j] = true;
    const h = heights[i][j];
    for (const [x, y] of dirs) {
      if (h <= (heights[i + x] ?? [])[j + y]) {
        dfs(i + x, j + y);
      }
    }
  };

  for (let i = 0; i < n; i++) {
    dfs(0, i);
  }
  for (let i = 0; i < m; i++) {
    dfs(i, 0);
  }
  isVis.forEach(v => v.fill(false));
  for (let i = 0; i < n; i++) {
    dfs(m - 1, i);
  }
  for (let i = 0; i < m; i++) {
    dfs(i, n - 1);
  }

  const res = [];
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (gird[i][j] === 2) {
        res.push([i, j]);
      }
    }
  }
  return res;
}