二维数组上的广度优先遍历 (BFS)

给定一个由整数组成的大小为M x N的矩阵，任务是使用广度优先搜索遍历打印矩阵元素。

例子：

Input: grid[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}}
Output: 1 2 5 3 6 9 4 7 10 13 8 11 14 12 15 16

Input: grid[][] = {{-1, 0, 0, 1}, {-1, -1, -2, -1}, {-1, -1, -1, -1}, {0, 0, 0, 0}}
Output: -1 0 -1 0 -1 -1 1 -2 -1 0 -1 -1 0 -1 0 0

处理方法：按照以下步骤解决问题：

1. 初始化方向向量dRow[] = {-1, 0, 1, 0}和dCol[] = {0, 1, 0, -1}和一对队列以存储矩阵单元的索引。
2. 从第一个单元格开始 BFS 遍历，即(0, 0) ，并将该单元格的索引加入队列。
3. 初始化一个布尔数组以标记矩阵的访问过的单元格。将单元格(0, 0)标记为已访问。
4. 声明一个函数isValid()来检查单元坐标是否有效，即位于给定矩阵的边界内并且是否未被访问。
5. 在队列不为空时迭代并执行以下操作：
   • 将出现在队列前面的单元出队并打印出来。
   • 移动到未访问的相邻单元格。
   • 将它们标记为已访问并将它们排入队列。

注意：方向向量用于以给定顺序遍历给定单元格的相邻单元格。例如(x, y)是一个单元格，它的相邻单元格(x – 1, y), (x, y + 1), (x + 1, y), (x, y – 1)需要遍历，那么它可以使用方向向量(-1, 0), (0, 1), (1, 0), (0, -1)按向上、向左、向下和向右的顺序完成。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
#define ROW 4
#define COL 4
 
// Direction vectors
int dRow[] = { -1, 0, 1, 0 };
int dCol[] = { 0, 1, 0, -1 };
 
// Function to check if a cell
// is be visited or not
bool isValid(bool vis[][COL],
             int row, int col)
{
    // If cell lies out of bounds
    if (row < 0 || col < 0
        || row >= ROW || col >= COL)
        return false;
 
    // If cell is already visited
    if (vis[row][col])
        return false;
 
    // Otherwise
    return true;
}
 
// Function to perform the BFS traversal
void BFS(int grid[][COL], bool vis[][COL],
         int row, int col)
{
    // Stores indices of the matrix cells
    queue > q;
 
    // Mark the starting cell as visited
    // and push it into the queue
    q.push({ row, col });
    vis[row][col] = true;
 
    // Iterate while the queue
    // is not empty
    while (!q.empty()) {
 
        pair cell = q.front();
        int x = cell.first;
        int y = cell.second;
 
        cout << grid[x][y] << " ";
 
        q.pop();
 
        // Go to the adjacent cells
        for (int i = 0; i < 4; i++) {
 
            int adjx = x + dRow[i];
            int adjy = y + dCol[i];
 
            if (isValid(vis, adjx, adjy)) {
                q.push({ adjx, adjy });
                vis[adjx][adjy] = true;
            }
        }
    }
}
 
// Driver Code
int main()
{
    // Given input matrix
    int grid[ROW][COL] = { { 1, 2, 3, 4 },
                           { 5, 6, 7, 8 },
                           { 9, 10, 11, 12 },
                           { 13, 14, 15, 16 } };
 
    // Declare the visited array
    bool vis[ROW][COL];
    memset(vis, false, sizeof vis);
 
    BFS(grid, vis, 0, 0);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
static class pair
{
    int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
 
static final int ROW = 4;
static final int COL = 4;
 
// Direction vectors
static int dRow[] = { -1, 0, 1, 0 };
static int dCol[] = { 0, 1, 0, -1 };
 
// Function to check if a cell
// is be visited or not
static boolean isValid(boolean vis[][],
                       int row, int col)
{
     
    // If cell lies out of bounds
    if (row < 0 || col < 0 ||
        row >= ROW || col >= COL)
        return false;
 
    // If cell is already visited
    if (vis[row][col])
        return false;
 
    // Otherwise
    return true;
}
 
// Function to perform the BFS traversal
static void BFS(int grid[][], boolean vis[][],
                int row, int col)
{
     
    // Stores indices of the matrix cells
    Queue q = new LinkedList<>();
 
    // Mark the starting cell as visited
    // and push it into the queue
    q.add(new pair(row, col));
    vis[row][col] = true;
 
    // Iterate while the queue
    // is not empty
    while (!q.isEmpty())
    {
        pair cell = q.peek();
        int x = cell.first;
        int y = cell.second;
 
        System.out.print(grid[x][y] + " ");
 
        q.remove();
 
        // Go to the adjacent cells
        for(int i = 0; i < 4; i++)
        {
            int adjx = x + dRow[i];
            int adjy = y + dCol[i];
 
            if (isValid(vis, adjx, adjy))
            {
                q.add(new pair(adjx, adjy));
                vis[adjx][adjy] = true;
            }
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given input matrix
    int grid[][] = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 9, 10, 11, 12 },
                     { 13, 14, 15, 16 } };
 
    // Declare the visited array
    boolean [][]vis = new boolean[ROW][COL];
 
    BFS(grid, vis, 0, 0);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
from collections import deque as queue
 
# Direction vectors
dRow = [ -1, 0, 1, 0]
dCol = [ 0, 1, 0, -1]
 
# Function to check if a cell
# is be visited or not
def isValid(vis, row, col):
   
    # If cell lies out of bounds
    if (row < 0 or col < 0 or row >= 4 or col >= 4):
        return False
 
    # If cell is already visited
    if (vis[row][col]):
        return False
 
    # Otherwise
    return True
 
# Function to perform the BFS traversal
def BFS(grid, vis, row, col):
   
    # Stores indices of the matrix cells
    q = queue()
 
    # Mark the starting cell as visited
    # and push it into the queue
    q.append(( row, col ))
    vis[row][col] = True
 
    # Iterate while the queue
    # is not empty
    while (len(q) > 0):
        cell = q.popleft()
        x = cell[0]
        y = cell[1]
        print(grid[x][y], end = " ")
 
        #q.pop()
 
        # Go to the adjacent cells
        for i in range(4):
            adjx = x + dRow[i]
            adjy = y + dCol[i]
            if (isValid(vis, adjx, adjy)):
                q.append((adjx, adjy))
                vis[adjx][adjy] = True
 
# Driver Code
if __name__ == '__main__':
   
    # Given input matrix
    grid= [ [ 1, 2, 3, 4 ],
           [ 5, 6, 7, 8 ],
           [ 9, 10, 11, 12 ],
           [ 13, 14, 15, 16 ] ]
 
    # Declare the visited array
    vis = [[ False for i in range(4)] for i in range(4)]
    # vis, False, sizeof vis)
 
    BFS(grid, vis, 0, 0)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
 
  class pair
  {
    public int first, second;
 
    public pair(int first, int second) 
    {
      this.first = first;
      this.second = second;
    }   
  }
  static readonly int ROW = 4;
  static readonly int COL = 4;
 
  // Direction vectors
  static int []dRow = { -1, 0, 1, 0 };
  static int []dCol = { 0, 1, 0, -1 };
 
  // Function to check if a cell
  // is be visited or not
  static bool isValid(bool [,]vis,
                      int row, int col)
  {
 
    // If cell lies out of bounds
    if (row < 0 || col < 0 ||
        row >= ROW || col >= COL)
      return false;
 
    // If cell is already visited
    if (vis[row,col])
      return false;
 
    // Otherwise
    return true;
  }
 
  // Function to perform the BFS traversal
  static void BFS(int [,]grid, bool [,]vis,
                  int row, int col)
  {
 
    // Stores indices of the matrix cells
    Queue q = new Queue();
 
    // Mark the starting cell as visited
    // and push it into the queue
    q.Enqueue(new pair(row, col));
    vis[row,col] = true;
 
    // Iterate while the queue
    // is not empty
    while (q.Count!=0)
    {
      pair cell = q.Peek();
      int x = cell.first;
      int y = cell.second;
      Console.Write(grid[x,y] + " ");
      q.Dequeue();
 
      // Go to the adjacent cells
      for(int i = 0; i < 4; i++)
      {
        int adjx = x + dRow[i];
        int adjy = y + dCol[i];
        if (isValid(vis, adjx, adjy))
        {
          q.Enqueue(new pair(adjx, adjy));
          vis[adjx,adjy] = true;
        }
      }
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
 
    // Given input matrix
    int [,]grid = { { 1, 2, 3, 4 },
                   { 5, 6, 7, 8 },
                   { 9, 10, 11, 12 },
                   { 13, 14, 15, 16 } };
 
    // Declare the visited array
    bool [,]vis = new bool[ROW,COL];
    BFS(grid, vis, 0, 0);
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program for the above approach
 
var ROW = 4;
var COL = 4;
 
// Direction vectors
var dRow = [-1, 0, 1, 0 ];
var dCol = [0, 1, 0, -1 ];
 
// Function to check if a cell
// is be visited or not
function isValid(vis, row, col)
{
    // If cell lies out of bounds
    if (row < 0 || col < 0
        || row >= ROW || col >= COL)
        return false;
 
    // If cell is already visited
    if (vis[row][col])
        return false;
 
    // Otherwise
    return true;
}
 
// Function to perform the BFS traversal
function BFS( grid, vis,row, col)
{
    // Stores indices of the matrix cells
    var q = [];
 
    // Mark the starting cell as visited
    // and push it into the queue
    q.push([row, col ]);
    vis[row][col] = true;
 
    // Iterate while the queue
    // is not empty
    while (q.length!=0) {
 
        var cell = q[0];
        var x = cell[0];
        var y = cell[1];
 
        document.write( grid[x][y] + " ");
 
        q.shift();
 
        // Go to the adjacent cells
        for (var i = 0; i < 4; i++) {
 
            var adjx = x + dRow[i];
            var adjy = y + dCol[i];
 
            if (isValid(vis, adjx, adjy)) {
                q.push([adjx, adjy ]);
                vis[adjx][adjy] = true;
            }
        }
    }
}
 
// Driver Code
// Given input matrix
var grid = [[1, 2, 3, 4 ],
                       [5, 6, 7, 8 ],
                       [9, 10, 11, 12 ],
                       [13, 14, 15, 16 ] ];
// Declare the visited array
var vis = Array.from(Array(ROW), ()=> Array(COL).fill(false));
BFS(grid, vis, 0, 0);
 
 

输出：

1 2 5 3 6 9 4 7 10 13 8 11 14 12 15 16

时间复杂度： O(N * M)
辅助空间： O(N * M)