代码随想录算法训练营第四十四天 | 99.岛屿数量 深搜 99.岛屿数量 广搜 100. 岛屿的最大面积

版本一的写法是 ：下一个节点是否能合法已经判断完了，传进dfs函数的就是合法节点。

版本二的写法是：不管节点是否合法，上来就dfs，然后在终止条件的地方进行判断，不合法再return。

我在之前回溯算法做过笔记，我更偏好版本一。

xy老让我联想到坐标，我就不用xy了。也可以叫row、col。

package main

import (
	"bufio"
	"fmt"
	"os"
)

var dir = [4][2]int{{0, 1}, {1, 0}, {-1, 0}, {0, -1}}

func main() {
	in := bufio.NewReader(os.Stdin)
	var n, m int
	fmt.Fscan(in, &n, &m)
	grid := make([][]int, n)
	visited := make([][]bool, n)
	for i := range grid {
		grid[i] = make([]int, m)
	}
	for i := range visited {
		visited[i] = make([]bool, m)
	}
	for i := range grid {
		for j := range grid[i] {
			fmt.Fscan(in, &grid[i][j])
		}
	}
	res := 0
	for i := range grid {
		for j := range grid[i] {
			if grid[i][j] == 1 && !visited[i][j] {
				res++
				visited[i][j] = true
				dfs(grid, visited, i, j)
			}
		}
	}
	fmt.Println(res)
}
func dfs(grid [][]int, visited [][]bool, i, j int) {
	for k := 0; k < 4; k++ {
		nextI := i + dir[k][0]
		nextJ := j + dir[k][1]
		if nextI < 0 || nextI >= len(grid) || nextJ < 0 || nextJ >= len(grid[0]) {
			continue
		}
		if grid[nextI][nextJ] == 1 && !visited[nextI][nextJ] {
			visited[nextI][nextJ] = true
			dfs(grid, visited, nextI, nextJ)
		}
	}
}

如果节点出队列再标记为已访问过，会导致相同的节点重复入队列，进而导致队列中会有大量的重复节点。

package main

import (
	"bufio"
	"fmt"
	"os"
)

var dir = [4][2]int{{0, 1}, {1, 0}, {-1, 0}, {0, -1}}

func main() {
	in := bufio.NewReader(os.Stdin)
	var n, m int
	fmt.Fscan(in, &n, &m)
	grid := make([][]int, n)
	visited := make([][]bool, n)
	for i := range grid {
		grid[i] = make([]int, m)
	}
	for i := range visited {
		visited[i] = make([]bool, m)
	}
	for i := range grid {
		for j := range grid[i] {
			fmt.Fscan(in, &grid[i][j])
		}
	}
	res := 0
	for i := range grid {
		for j := range grid[i] {
			if grid[i][j] == 1 && !visited[i][j] {
				res++
				visited[i][j] = true
				bfs(grid, visited, i, j)
			}
		}
	}
	fmt.Println(res)
}

type Pair struct {
	i, j int
}

func bfs(grid [][]int, visited [][]bool, row, col int) {
	q := make([]Pair, 0)
	q = append(q, Pair{row, col})
	visited[row][col] = true

	for len(q) != 0 {
		cur := q[0]
		q = q[1:]
		for k := 0; k < 4; k++ {
			nextI := cur.i + dir[k][0]
			nextJ := cur.j + dir[k][1]
			if nextI < 0 || nextI >= len(grid) || nextJ < 0 || nextJ >= len(grid[0]) {
				continue
			}
			if grid[nextI][nextJ] == 1 && !visited[nextI][nextJ] {
				q = append(q, Pair{nextI, nextJ})
				visited[nextI][nextJ] = true
			}
		}
	}
}

easy

package main

import (
	"bufio"
	"fmt"
	"os"
)

var dir = [4][2]int{{0, 1}, {1, 0}, {-1, 0}, {0, -1}}
var count int

func main() {
	in := bufio.NewReader(os.Stdin)
	var n, m int
	fmt.Fscan(in, &n, &m)
	grid := make([][]int, n)
	visited := make([][]bool, n)
	for i := range grid {
		grid[i] = make([]int, m)
	}
	for i := range visited {
		visited[i] = make([]bool, m)
	}
	for i := range grid {
		for j := range grid[i] {
			fmt.Fscan(in, &grid[i][j])
		}
	}
	res := 0
	for i := range grid {
		for j := range grid[i] {
			if grid[i][j] == 1 && !visited[i][j] {
				visited[i][j] = true
				count = 1
				dfs(grid, visited, i, j)
				if count > res {
					res = count
				}
			}
		}
	}
	fmt.Println(res)
}
func dfs(grid [][]int, visited [][]bool, i, j int) {
	for k := 0; k < 4; k++ {
		nextI := i + dir[k][0]
		nextJ := j + dir[k][1]
		if nextI < 0 || nextI >= len(grid) || nextJ < 0 || nextJ >= len(grid[0]) {
			continue
		}
		if grid[nextI][nextJ] == 1 && !visited[nextI][nextJ] {
			visited[nextI][nextJ] = true
			count++
			dfs(grid, visited, nextI, nextJ)
		}
	}
}