417. Pacific Atlantic Water Flow

Medium
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:
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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] <= 10^5

解題

這個解法不是對每個點做 DFS,讓水從高處往下流、把能流進兩大洋的加入答案。
而是從邊邊往內找山脊!一旦跨過山脊發現下坡就停止。
func pacificAtlantic(heights [][]int) [][]int {
res := make([][]int, 0)
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// 初始化
pacific := make([][]bool, 0)
atlantic := make([][]bool, 0)
for i := 0; i < len(heights); i++ {
pacific = append(pacific, make([]bool, len(heights[0])))
atlantic = append(atlantic, make([]bool, len(heights[0])))
}
​
for i := 0; i < len(heights); i++ {
dfs(heights, pacific, math.MinInt32, i, 0); // å·¦
dfs(heights, atlantic, math.MinInt32, i, len(heights[0]) - 1); // 右
}
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for i := 0; i < len(heights[0]); i++ {
dfs(heights, pacific, math.MinInt32, 0, i); // 最上排
dfs(heights, atlantic, math.MinInt32, len(heights) - 1, i); //最下排
}
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for i := 0; i < len(heights); i++ {
for j := 0; j < len(heights[0]); j++ {
if pacific[i][j] && atlantic[i][j] {
res = append(res, []int{i, j})
}
}
}
​
return res
}
​
func dfs(heights [][]int, sea [][]bool, pre, i, j int) {
// 超過邊界 / 遇到下坡/ 已經找過 就停止
if i < 0 || j < 0 || i >= len(heights) || j >= len(heights[0]) || sea[i][j] || heights[i][j] < pre {
return
}
​
sea[i][j] = true
dfs(heights, sea, heights[i][j], i + 1, j)
dfs(heights, sea, heights[i][j], i - 1, j)
dfs(heights, sea, heights[i][j], i, j + 1)
dfs(heights, sea, heights[i][j], i, j - 1)
}
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