README.md

200. Number of Islands

Problem Statement

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.

 

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'.

Solutions

Go

package main

func numIslands(grid [][]byte) int {
	result := 0

	for i := 0; i < len(grid); i++ {
		for j := 0; j < len(grid[i]); j++ {
			if grid[i][j] == '1' {
				dfs(grid, i, j)
				result++
			}
		}
	}

	return result
}

func dfs(grid [][]byte, row, col int) {
	// return if it is out of bound horizontally
	if row < 0 || row >= len(grid) {
		return
	}

	// return if it is out of bound vertically
	if col < 0 || col >= len(grid[row]) {
		return
	}

	// return if it is not an island
	if grid[row][col] == '0' {
		return
	}

	// mark the current cell as visited
	grid[row][col] = '0'

	dfs(grid, row-1, col)
	dfs(grid, row+1, col)

	dfs(grid, row, col-1)
	dfs(grid, row, col+1)
}

...