動機

要用dfs(+dp)要注意會不會重複走

Problem

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]]

Example 2:

Input: heights = [[2,1],[1,2]]Output: [[0,0],[0,1],[1,0],[1,1]]

 

Constraints:

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

Sol

當初腦袋撞到，一直想用dp去做，但不是不能用dp做，不過用記憶法不好做 要

因為要處理重複走的問題

如果dfs不行，我們還有bfs!!

class Solution:
    # 如果要用dfs(+dp)要注意會不會重複走，如果會就很麻煩
    # 同樣是搜尋所有可能性可以用bfs，尤其是起點或終點已經確定的更是如此
    def pacificAtlantic(self, hs: List[List[int]]) -> List[List[int]]:
        def legal(x,y):
            return 0 <= x < len(hs) and 0 <= y < len(hs[x])
        def bfs(starts):
            q,vis = deque(starts),set(starts)
            while q:
                (i,j) = q.popleft()
                dirs = [(i+1,j),(i,j+1),(i-1,j),(i,j-1)]
                dirs = [(x,y) for (x,y) in dirs if legal(x,y) and (x,y) not in vis and hs[i][j] <= hs[x][y]]
                for xy in dirs:
                    q.append(xy)
                    vis.add(xy)
            return vis
        if hs:
            p = [(0,j) for j in range(len(hs[0]))] + [(i,0) for i in range(len(hs))]
            a = [(len(hs)-1,j) for j in range(len(hs[0]))] + [(i,len(hs[i])-1) for i in range(len(hs))]
            return bfs(p) & bfs(a)
        else:
            return []