golang_number_of_islands

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Examples

Example 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is '0' or '1'.

解析

給定一個 2 D 二元矩陣 grid ， 每個元素 grid[r][c] 只有 1 或是 0 兩種值

1 代表是陸地，0 代表是 水

如果水平或是垂直相鄰位置都是 1 代表 cell屬於同一個 island

題目實作一個演算法從 grid 找出總共有幾個 island

直覺去思考，可以發現要找出有幾個 island 代表要去找有幾個相連的區塊

而要找出相連的區塊會透過 BFS 從該區塊相鄰為 1 的 cell 開始找尋直到遇到 0 或是 邊界

如下圖

會發現需要紀錄每個拜訪過的 grid，可以使用 hashSet

然後逐步每個 cell 檢查

程式碼

package sol

type Pair struct {
	row, col int
}

func numIslands(grid [][]byte) int {
	ROW := len(grid)
	COL := len(grid[0])
	visit := make([][]bool, ROW)
	for row := range visit {
		visit[row] = make([]bool, COL)
		for col := range visit[row] {
			visit[row][col] = false
		}
	}
	island := 0
	directions := []Pair{{row: -1, col: 0}, {row: 1, col: 0}, {row: 0, col: -1}, {row: 0, col: 1}}
	var bfs = func(row int, col int) {
		queue := []Pair{{row: row, col: col}}
		visit[row][col] = true
		for len(queue) != 0 {
			top := queue[0]
			queue = queue[1:]
			for _, direction := range directions {
				shifted_row := top.row + direction.row
				shifted_col := top.col + direction.col
				if shifted_row < 0 || shifted_row >= ROW ||
					shifted_col < 0 || shifted_col >= COL ||
					visit[shifted_row][shifted_col] || grid[shifted_row][shifted_col] == '0' {
					continue
				}
				visit[shifted_row][shifted_col] = true
				queue = append(queue, Pair{row: shifted_row, col: shifted_col})
			}
		}
	}
	for row := 0; row < ROW; row++ {
		for col := 0; col < COL; col++ {
			if grid[row][col] == '1' && !visit[row][col] {
				bfs(row, col)
				island++
			}
		}
	}
	return island
}

困難點

1. 需要理解找出連結在一起 cell 要使用 BFS
2. 需要理解 BFS 實作

Solve Point

• Understand what problem to solve
• Analysis Complexity