1992. Find All Groups of Farmland
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# 1992. Find All Groups of Farmland

Leetcode #1992

## 題目描述

You are given a 0-indexed m x n binary matrix land where a 0 represents a hectare of forested land and a 1 represents a hectare of farmland.

To keep the land organized, there are designated rectangular areas of hectares that consist entirely of farmland. These rectangular areas are called groups. No two groups are adjacent, meaning farmland in one group is not four-directionally adjacent to another farmland in a different group.

land can be represented by a coordinate system where the top left corner of land is (0, 0) and the bottom right corner of land is (m-1, n-1). Find the coordinates of the top left and bottom right corner of each group of farmland. A group of farmland with a top left corner at (r1, c1) and a bottom right corner at (r2, c2) is represented by the 4-length array [r1, c1, r2, c2].

Return a 2D array containing the 4-length arrays described above for each group of farmland in land. If there are no groups of farmland, return an empty array. You may return the answer in any order.

### 題目限制

• m == land.length
• n == land[i].length
• 1 <= m, n <= 300
• land consists of only 0’s and 1’s.
• Groups of farmland are rectangular in shape.

## 具體測資

### Case 1

1 2 3 4 5 Input: land = [[1,0,0],[0,1,1],[0,1,1]] Output: [[0,0,0,0],[1,1,2,2]] Explanation: The first group has a top left corner at land[0][0] and a bottom right corner at land[0][0]. The second group has a top left corner at land[1][1] and a bottom right corner at land[2][2]. 

### Case 2

1 2 3 4 Input: land = [[1,1],[1,1]] Output: [[0,0,1,1]] Explanation: The first group has a top left corner at land[0][0] and a bottom right corner at land[1][1]. 

### Case 3

1 2 3 4 Input: land = [[0]] Output: [] Explanation: There are no groups of farmland. 

## 解題想法

### Pseudo Code

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 GIVEN: land is a 2D array. SET m is len(land). SET n is len(land[0]). SET visited_map is a 2D (mxn) array where all elements are False. SET ret is a empty list. FUNC dfs(i, j) do IF i >= m or i < 0 or j >= n or j < 0: RETURN (0, 0) ENDIF IF visited_map[i][j]: RETURN (0, 0) ENDIF IF land[i][j] == 0: RETURN (0, 0) ENDIF visited_map[i][j] = True RETURN max((i, j), dfs(i+1, j), dfs(i, j+1)) ENDFUNC FOR i (0 to m) do FOR j (0 to n) do IF not visited_map[i][j] do IF land[i][j] == 1 do (p, q) <- dfs(i, j) ret <- ret + [i, j, p, q] ELSE visited_map[i][j] <- True ENDIF ENDIF ENDFOR ENDFOR RETURN ret 

### 複雜度分析

• 時間複雜度: O(m*n)

• 空間複雜度: O(1) (不考慮回傳時所創建的矩陣)

## 實際解題

### Python

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 class Solution: def findFarmland(self, land: List[List[int]]) -> List[List[int]]: m = len(land) n = len(land[0]) visited_map = [[False] * n for _ in range(m)] ## 以 (i, j) 為矩形左上座標之視角，遍歷出矩形右下座標 def dfs(i: int, j: int): if i >= m or i < 0 or j >= n or j < 0: return (0, 0) if visited_map[i][j]: return (0, 0) if land[i][j] == 0: return (0, 0) visited_map[i][j] = True return max((i, j), dfs(i+1, j), dfs(i, j+1)) ret = list() for i in range(m): for j in range(n): ## 如果 (i, j) 還未遍歷過 if not visited_map[i][j]: ## 此時該點必為矩形左上座標 if land[i][j] == 1: ## 找尋與該點配對之右下座標 (p, q) = dfs(i, j) ret.append([i, j, p, q]) else: visited_map[i][j] = True return ret