给定大小为M x N的整数矩阵，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
方法：请按照以下步骤解决问题：
- 初始化方向向量dRow [] = { -1，0，1，0 }和dCol [] = {0，1，0，-1}和成对的队列以存储矩阵单元的索引。
- 从第一个单元(即(0，0))开始BFS遍历，并将该单元的索引排入队列。
- 初始化一个布尔数组以标记矩阵的访问单元。将单元格(0，0)标记为已访问。
- 声明一个函数isValid()，以检查单元格坐标是否有效，即是否位于给定Matrix的边界内，并且未被访问。
- 在队列不为空时进行迭代并执行以下操作：
- 使队列最前面的单元格出队并进行打印。
- 移至其未访问的相邻单元格。
- 将他们标记为已访问并将他们排队。
注意：方向向量用于按给定顺序遍历给定单元格的相邻单元格，例如(x，y)是其相邻单元格(x – 1，y)，(x，y + 1)，( x + 1，y)，(x，y – 1)需要遍历，然后可以使用方向矢量(-1，0)，(0，1)，(1，0)，(0，- 1)按上，左，下和右顺序。
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 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)
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