查找岛屿数量 |设置 1（使用 DFS）
给定一个布尔二维矩阵，找出岛屿的数量。一组连接的 1 形成一个岛。例如，下面的矩阵包含 5 个岛屿
例子：
Input : mat[][] = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}}
Output : 5这是标准问题的变体：“计算无向图中的连通分量的数量”。
在我们解决问题之前，让我们了解什么是连接组件。无向图的连通分量是一个子图，其中每两个顶点通过一条路径相互连接，并且不与子图之外的其他顶点相连。
例如，下图具有三个连通分量。
_0.png)
所有顶点相互连接的图只有一个连通分量，由整个图组成。这种只有一个连通分量的图称为强连通图。
通过在每个组件上应用 DFS() 可以轻松解决该问题。在每个 DFS() 调用中，都会访问一个组件或一个子图。我们将在下一个未访问的组件上调用 DFS。对 DFS() 的调用次数给出了连接组件的数量。也可以使用 BFS。
什么是岛？
一组连接的 1 形成一个岛。例如，下面的矩阵包含 4 个岛屿
_1.png)
二维矩阵中的一个单元可以连接到 8 个邻居。因此，与标准 DFS() 不同的是，我们递归调用所有相邻顶点，这里我们只能递归调用 8 个邻居。我们跟踪访问过的 1，以便它们不再被访问。
C++
// C++ Program to count islands in boolean 2D matrix
#include
using namespace std;
#define ROW 5
#define COL 5
// A function to check if a given
// cell (row, col) can be included in DFS
int isSafe(int M[][COL], int row, int col,
bool visited[][COL])
{
// row number is in range, column
// number is in range and value is 1
// and not yet visited
return (row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row][col] && !visited[row][col]);
}
// A utility function to do DFS for a
// 2D boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(int M[][COL], int row, int col,
bool visited[][COL])
{
// These arrays are used to get
// row and column numbers of 8
// neighbours of a given cell
static int rowNbr[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
static int colNbr[] = { -1, 0, 1, -1, 1, -1, 0, 1 };
// Mark this cell as visited
visited[row][col] = true;
// Recur for all connected neighbours
for (int k = 0; k < 8; ++k)
if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited))
DFS(M, row + rowNbr[k], col + colNbr[k], visited);
}
// The main function that returns
// count of islands in a given boolean
// 2D matrix
int countIslands(int M[][COL])
{
// Make a bool array to mark visited cells.
// Initially all cells are unvisited
bool visited[ROW][COL];
memset(visited, 0, sizeof(visited));
// Initialize count as 0 and
// traverse through the all cells of
// given matrix
int count = 0;
for (int i = 0; i < ROW; ++i)
for (int j = 0; j < COL; ++j)
// If a cell with value 1 is not
if (M[i][j] && !visited[i][j]) {
// visited yet, then new island found
// Visit all cells in this island.
DFS(M, i, j, visited);
// and increment island count
++count;
}
return count;
}
// Driver code
int main()
{
int M[][COL] = { { 1, 1, 0, 0, 0 },
{ 0, 1, 0, 0, 1 },
{ 1, 0, 0, 1, 1 },
{ 0, 0, 0, 0, 0 },
{ 1, 0, 1, 0, 1 } };
cout << "Number of islands is: " << countIslands(M);
return 0;
}
// This is code is contributed by rathbhupendra C
// Program to count islands in boolean 2D matrix
#include
#include
#include
#define ROW 5
#define COL 5
// A function to check if a given cell (row, col) can be included in DFS
int isSafe(int M[][COL], int row, int col, bool visited[][COL])
{
// row number is in range, column number is in range and value is 1
// and not yet visited
return (row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row][col] && !visited[row][col]);
}
// A utility function to do DFS for a 2D boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(int M[][COL], int row, int col, bool visited[][COL])
{
// These arrays are used to get row and column numbers of 8 neighbours
// of a given cell
static int rowNbr[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
static int colNbr[] = { -1, 0, 1, -1, 1, -1, 0, 1 };
// Mark this cell as visited
visited[row][col] = true;
// Recur for all connected neighbours
for (int k = 0; k < 8; ++k)
if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited))
DFS(M, row + rowNbr[k], col + colNbr[k], visited);
}
// The main function that returns count of islands in a given boolean
// 2D matrix
int countIslands(int M[][COL])
{
// Make a bool array to mark visited cells.
// Initially all cells are unvisited
bool visited[ROW][COL];
memset(visited, 0, sizeof(visited));
// Initialize count as 0 and traverse through the all cells of
// given matrix
int count = 0;
for (int i = 0; i < ROW; ++i)
for (int j = 0; j < COL; ++j)
if (M[i][j] && !visited[i][j]) // If a cell with value 1 is not
{ // visited yet, then new island found
DFS(M, i, j, visited); // Visit all cells in this island.
++count; // and increment island count
}
return count;
}
// Driver program to test above function
int main()
{
int M[][COL] = { { 1, 1, 0, 0, 0 },
{ 0, 1, 0, 0, 1 },
{ 1, 0, 0, 1, 1 },
{ 0, 0, 0, 0, 0 },
{ 1, 0, 1, 0, 1 } };
printf("Number of islands is: %d\n", countIslands(M));
return 0;
} Java
// Java program to count islands in boolean 2D matrix
import java.util.*;
import java.lang.*;
import java.io.*;
class Islands {
// No of rows and columns
static final int ROW = 5, COL = 5;
// A function to check if a given cell (row, col) can
// be included in DFS
boolean isSafe(int M[][], int row, int col,
boolean visited[][])
{
// row number is in range, column number is in range
// and value is 1 and not yet visited
return (row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row][col] == 1 && !visited[row][col]);
}
// A utility function to do DFS for a 2D boolean matrix.
// It only considers the 8 neighbors as adjacent vertices
void DFS(int M[][], int row, int col, boolean visited[][])
{
// These arrays are used to get row and column numbers
// of 8 neighbors of a given cell
int rowNbr[] = new int[] { -1, -1, -1, 0, 0, 1, 1, 1 };
int colNbr[] = new int[] { -1, 0, 1, -1, 1, -1, 0, 1 };
// Mark this cell as visited
visited[row][col] = true;
// Recur for all connected neighbours
for (int k = 0; k < 8; ++k)
if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited))
DFS(M, row + rowNbr[k], col + colNbr[k], visited);
}
// The main function that returns count of islands in a given
// boolean 2D matrix
int countIslands(int M[][])
{
// Make a bool array to mark visited cells.
// Initially all cells are unvisited
boolean visited[][] = new boolean[ROW][COL];
// Initialize count as 0 and traverse through the all cells
// of given matrix
int count = 0;
for (int i = 0; i < ROW; ++i)
for (int j = 0; j < COL; ++j)
if (M[i][j] == 1 && !visited[i][j]) // If a cell with
{ // value 1 is not
// visited yet, then new island found, Visit all
// cells in this island and increment island count
DFS(M, i, j, visited);
++count;
}
return count;
}
// Driver method
public static void main(String[] args) throws java.lang.Exception
{
int M[][] = new int[][] { { 1, 1, 0, 0, 0 },
{ 0, 1, 0, 0, 1 },
{ 1, 0, 0, 1, 1 },
{ 0, 0, 0, 0, 0 },
{ 1, 0, 1, 0, 1 } };
Islands I = new Islands();
System.out.println("Number of islands is: " + I.countIslands(M));
}
} // Contributed by Aakash HasijaPython3
# Program to count islands in boolean 2D matrix
class Graph:
def __init__(self, row, col, g):
self.ROW = row
self.COL = col
self.graph = g
# A function to check if a given cell
# (row, col) can be included in DFS
def isSafe(self, i, j, visited):
# row number is in range, column number
# is in range and value is 1
# and not yet visited
return (i >= 0 and i < self.ROW and
j >= 0 and j < self.COL and
not visited[i][j] and self.graph[i][j])
# A utility function to do DFS for a 2D
# boolean matrix. It only considers
# the 8 neighbours as adjacent vertices
def DFS(self, i, j, visited):
# These arrays are used to get row and
# column numbers of 8 neighbours
# of a given cell
rowNbr = [-1, -1, -1, 0, 0, 1, 1, 1];
colNbr = [-1, 0, 1, -1, 1, -1, 0, 1];
# Mark this cell as visited
visited[i][j] = True
# Recur for all connected neighbours
for k in range(8):
if self.isSafe(i + rowNbr[k], j + colNbr[k], visited):
self.DFS(i + rowNbr[k], j + colNbr[k], visited)
# The main function that returns
# count of islands in a given boolean
# 2D matrix
def countIslands(self):
# Make a bool array to mark visited cells.
# Initially all cells are unvisited
visited = [[False for j in range(self.COL)]for i in range(self.ROW)]
# Initialize count as 0 and traverse
# through the all cells of
# given matrix
count = 0
for i in range(self.ROW):
for j in range(self.COL):
# If a cell with value 1 is not visited yet,
# then new island found
if visited[i][j] == False and self.graph[i][j] == 1:
# Visit all cells in this island
# and increment island count
self.DFS(i, j, visited)
count += 1
return count
graph = [[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]]
row = len(graph)
col = len(graph[0])
g = Graph(row, col, graph)
print ("Number of islands is:")
print (g.countIslands())
# This code is contributed by Neelam YadavC#
// C# program to count
// islands in boolean
// 2D matrix
using System;
class GFG {
// No of rows
// and columns
static int ROW = 5, COL = 5;
// A function to check if
// a given cell (row, col)
// can be included in DFS
static bool isSafe(int[, ] M, int row,
int col, bool[, ] visited)
{
// row number is in range,
// column number is in range
// and value is 1 and not
// yet visited
return (row >= 0) && (row < ROW) && (col >= 0) && (col < COL) && (M[row, col] == 1 && !visited[row, col]);
}
// A utility function to do
// DFS for a 2D boolean matrix.
// It only considers the 8
// neighbors as adjacent vertices
static void DFS(int[, ] M, int row,
int col, bool[, ] visited)
{
// These arrays are used to
// get row and column numbers
// of 8 neighbors of a given cell
int[] rowNbr = new int[] { -1, -1, -1, 0,
0, 1, 1, 1 };
int[] colNbr = new int[] { -1, 0, 1, -1,
1, -1, 0, 1 };
// Mark this cell
// as visited
visited[row, col] = true;
// Recur for all
// connected neighbours
for (int k = 0; k < 8; ++k)
if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited))
DFS(M, row + rowNbr[k],
col + colNbr[k], visited);
}
// The main function that
// returns count of islands
// in a given boolean 2D matrix
static int countIslands(int[, ] M)
{
// Make a bool array to
// mark visited cells.
// Initially all cells
// are unvisited
bool[, ] visited = new bool[ROW, COL];
// Initialize count as 0 and
// traverse through the all
// cells of given matrix
int count = 0;
for (int i = 0; i < ROW; ++i)
for (int j = 0; j < COL; ++j)
if (M[i, j] == 1 && !visited[i, j]) {
// If a cell with value 1 is not
// visited yet, then new island
// found, Visit all cells in this
// island and increment island count
DFS(M, i, j, visited);
++count;
}
return count;
}
// Driver Code
public static void Main()
{
int[, ] M = new int[, ] { { 1, 1, 0, 0, 0 },
{ 0, 1, 0, 0, 1 },
{ 1, 0, 0, 1, 1 },
{ 0, 0, 0, 0, 0 },
{ 1, 0, 1, 0, 1 } };
Console.Write("Number of islands is: " + countIslands(M));
}
}
// This code is contributed
// by shiv_bhakt.PHP
= 0) && ($row < $ROW) &&
($col >= 0) && ($col < $COL) &&
($M[$row][$col] &&
!isset($visited[$row][$col]));
}
// A utility function to do DFS
// for a 2D boolean matrix. It
// only considers the 8 neighbours
// as adjacent vertices
function DFS(&$M, $row, $col,
&$visited)
{
// These arrays are used to
// get row and column numbers
// of 8 neighbours of a given cell
$rowNbr = array(-1, -1, -1, 0,
0, 1, 1, 1);
$colNbr = array(-1, 0, 1, -1,
1, -1, 0, 1);
// Mark this cell as visited
$visited[$row][$col] = true;
// Recur for all
// connected neighbours
for ($k = 0; $k < 8; ++$k)
if (isSafe($M, $row + $rowNbr[$k],
$col + $colNbr[$k], $visited))
DFS($M, $row + $rowNbr[$k],
$col + $colNbr[$k], $visited);
}
// The main function that returns
// count of islands in a given
// boolean 2D matrix
function countIslands(&$M)
{
global $ROW, $COL;
// Make a bool array to
// mark visited cells.
// Initially all cells
// are unvisited
$visited = array(array());
// Initialize count as 0 and
// traverse through the all
// cells of given matrix
$count = 0;
for ($i = 0; $i < $ROW; ++$i)
for ($j = 0; $j < $COL; ++$j)
if ($M[$i][$j] &&
!isset($visited[$i][$j])) // If a cell with value 1
{ // is not visited yet,
DFS($M, $i, $j, $visited); // then new island found
++$count; // Visit all cells in this
} // island and increment
// island count.
return $count;
}
// Driver Code
$M = array(array(1, 1, 0, 0, 0),
array(0, 1, 0, 0, 1),
array(1, 0, 0, 1, 1),
array(0, 0, 0, 0, 0),
array(1, 0, 1, 0, 1));
echo "Number of islands is: ",
countIslands($M);
// This code is contributed
// by ChitraNayal
?>Javascript
C++
// C++Program to count islands in boolean 2D matrix
#include
using namespace std;
// A utility function to do DFS for a 2D
// boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(vector> &M, int i, int j, int ROW,
int COL)
{
//Base condition
//if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL- or if M[i][j] != 1 then we will simply return
if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
{
return;
}
if (M[i][j] == 1)
{
M[i][j] = 0;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
int countIslands(vector> &M)
{
int ROW = M.size();
int COL = M[0].size();
int count = 0;
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (M[i][j] == 1)
{
M[i][j] = 0;
count++;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
}
return count;
}
// Driver Code
int main()
{
vector> M = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}};
cout << "Number of islands is: " << countIslands(M);
return 0;
}
// This code is contributed by ajaymakvana. Java
// Java Program to count islands in boolean 2D matrix
import java.util.*;
public class Main
{
// A utility function to do DFS for a 2D
// boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
static void DFS(int[][] M, int i, int j, int ROW, int COL)
{
// Base condition
// if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL- or if M[i][j] != 1 then we will simply return
if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
{
return;
}
if (M[i][j] == 1)
{
M[i][j] = 0;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
static int countIslands(int[][] M)
{
int ROW = M.length;
int COL = M[0].length;
int count = 0;
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (M[i][j] == 1)
{
M[i][j] = 0;
count++;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
}
return count;
}
// Driver code
public static void main(String[] args) {
int[][] M = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}};
System.out.print("Number of islands is: " + countIslands(M));
}
}
// This code is contributed by suresh07.Python3
# Program to count islands in boolean 2D matrix
class Graph:
def __init__(self, row, col, graph):
self.ROW = row
self.COL = col
self.graph = graph
# A utility function to do DFS for a 2D
# boolean matrix. It only considers
# the 8 neighbours as adjacent vertices
def DFS(self, i, j):
if i < 0 or i >= len(self.graph) or j < 0 or j >= len(self.graph[0]) or self.graph[i][j] != 1:
return
# mark it as visited
self.graph[i][j] = -1
# Recur for 8 neighbours
self.DFS(i - 1, j - 1)
self.DFS(i - 1, j)
self.DFS(i - 1, j + 1)
self.DFS(i, j - 1)
self.DFS(i, j + 1)
self.DFS(i + 1, j - 1)
self.DFS(i + 1, j)
self.DFS(i + 1, j + 1)
# The main function that returns
# count of islands in a given boolean
# 2D matrix
def countIslands(self):
# Initialize count as 0 and traverse
# through the all cells of
# given matrix
count = 0
for i in range(self.ROW):
for j in range(self.COL):
# If a cell with value 1 is not visited yet,
# then new island found
if self.graph[i][j] == 1:
# Visit all cells in this island
# and increment island count
self.DFS(i, j)
count += 1
return count
graph = [
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]
]
row = len(graph)
col = len(graph[0])
g = Graph(row, col, graph)
print("Number of islands is:", g.countIslands())
# This code is contributed by Shivam ShreyC#
// C# Program to count islands in boolean 2D matrix
using System;
using System.Collections.Generic;
class GFG {
// A utility function to do DFS for a 2D
// boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
static void DFS(int[,] M, int i, int j, int ROW, int COL)
{
// Base condition
// if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL- or if M[i][j] != 1 then we will simply return
if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i,j] != 1)
{
return;
}
if (M[i,j] == 1)
{
M[i,j] = 0;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
static int countIslands(int[,] M)
{
int ROW = M.GetLength(0);
int COL = M.GetLength(1);
int count = 0;
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (M[i,j] == 1)
{
M[i,j] = 0;
count++;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
}
return count;
}
// Driver code
static void Main() {
int[,] M = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}};
Console.Write("Number of islands is: " + countIslands(M));
}
}
// This code is contributed by decode2207.Javascript
输出
Number of islands is: 5时间复杂度： O(ROW x COL)
不创建访问矩阵的替代解决方案：
C++
// C++Program to count islands in boolean 2D matrix
#include
using namespace std;
// A utility function to do DFS for a 2D
// boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(vector> &M, int i, int j, int ROW,
int COL)
{
//Base condition
//if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL- or if M[i][j] != 1 then we will simply return
if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
{
return;
}
if (M[i][j] == 1)
{
M[i][j] = 0;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
int countIslands(vector> &M)
{
int ROW = M.size();
int COL = M[0].size();
int count = 0;
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (M[i][j] == 1)
{
M[i][j] = 0;
count++;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
}
return count;
}
// Driver Code
int main()
{
vector> M = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}};
cout << "Number of islands is: " << countIslands(M);
return 0;
}
// This code is contributed by ajaymakvana.
Java
// Java Program to count islands in boolean 2D matrix
import java.util.*;
public class Main
{
// A utility function to do DFS for a 2D
// boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
static void DFS(int[][] M, int i, int j, int ROW, int COL)
{
// Base condition
// if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL- or if M[i][j] != 1 then we will simply return
if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i][j] != 1)
{
return;
}
if (M[i][j] == 1)
{
M[i][j] = 0;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
static int countIslands(int[][] M)
{
int ROW = M.length;
int COL = M[0].length;
int count = 0;
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (M[i][j] == 1)
{
M[i][j] = 0;
count++;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
}
return count;
}
// Driver code
public static void main(String[] args) {
int[][] M = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}};
System.out.print("Number of islands is: " + countIslands(M));
}
}
// This code is contributed by suresh07.
Python3
# Program to count islands in boolean 2D matrix
class Graph:
def __init__(self, row, col, graph):
self.ROW = row
self.COL = col
self.graph = graph
# A utility function to do DFS for a 2D
# boolean matrix. It only considers
# the 8 neighbours as adjacent vertices
def DFS(self, i, j):
if i < 0 or i >= len(self.graph) or j < 0 or j >= len(self.graph[0]) or self.graph[i][j] != 1:
return
# mark it as visited
self.graph[i][j] = -1
# Recur for 8 neighbours
self.DFS(i - 1, j - 1)
self.DFS(i - 1, j)
self.DFS(i - 1, j + 1)
self.DFS(i, j - 1)
self.DFS(i, j + 1)
self.DFS(i + 1, j - 1)
self.DFS(i + 1, j)
self.DFS(i + 1, j + 1)
# The main function that returns
# count of islands in a given boolean
# 2D matrix
def countIslands(self):
# Initialize count as 0 and traverse
# through the all cells of
# given matrix
count = 0
for i in range(self.ROW):
for j in range(self.COL):
# If a cell with value 1 is not visited yet,
# then new island found
if self.graph[i][j] == 1:
# Visit all cells in this island
# and increment island count
self.DFS(i, j)
count += 1
return count
graph = [
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[1, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[1, 0, 1, 0, 1]
]
row = len(graph)
col = len(graph[0])
g = Graph(row, col, graph)
print("Number of islands is:", g.countIslands())
# This code is contributed by Shivam Shrey
C＃
// C# Program to count islands in boolean 2D matrix
using System;
using System.Collections.Generic;
class GFG {
// A utility function to do DFS for a 2D
// boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
static void DFS(int[,] M, int i, int j, int ROW, int COL)
{
// Base condition
// if i less than 0 or j less than 0 or i greater than ROW-1 or j greater than COL- or if M[i][j] != 1 then we will simply return
if (i < 0 || j < 0 || i > (ROW - 1) || j > (COL - 1) || M[i,j] != 1)
{
return;
}
if (M[i,j] == 1)
{
M[i,j] = 0;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
static int countIslands(int[,] M)
{
int ROW = M.GetLength(0);
int COL = M.GetLength(1);
int count = 0;
for (int i = 0; i < ROW; i++)
{
for (int j = 0; j < COL; j++)
{
if (M[i,j] == 1)
{
M[i,j] = 0;
count++;
DFS(M, i + 1, j, ROW, COL); //right side traversal
DFS(M, i - 1, j, ROW, COL); //left side traversal
DFS(M, i, j + 1, ROW, COL); //upward side traversal
DFS(M, i, j - 1, ROW, COL); //downward side traversal
DFS(M, i + 1, j + 1, ROW, COL); //upward-right side traversal
DFS(M, i - 1, j - 1, ROW, COL); //downward-left side traversal
DFS(M, i + 1, j - 1, ROW, COL); //downward-right side traversal
DFS(M, i - 1, j + 1, ROW, COL); //upward-left side traversal
}
}
}
return count;
}
// Driver code
static void Main() {
int[,] M = {{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}};
Console.Write("Number of islands is: " + countIslands(M));
}
}
// This code is contributed by decode2207.
Javascript
输出
Number of islands is: 5时间复杂度： O(ROW x COL)