给定一个 2D 屏幕arr[][] ，其中每个arr[i][j]是一个表示该像素颜色的整数，还给定了像素(X, Y) 的位置和颜色C ，任务是替换给定像素的颜色以及具有给定颜色的所有相邻相同颜色的像素。
例子：
Input: arr[][] = {
{1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 0, 0},
{1, 0, 0, 1, 1, 0, 1, 1},
{1, 2, 2, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 2, 2, 0},
{1, 1, 1, 1, 1, 2, 1, 1},
{1, 1, 1, 1, 1, 2, 2, 1}}
X = 4, Y = 4, C = 3
Output:
1 1 1 1 1 1 1 1
1 1 1 1 1 1 0 0
1 0 0 1 1 0 1 1
1 3 3 3 3 0 1 0
1 1 1 3 3 0 1 0
1 1 1 3 3 3 3 0
1 1 1 1 1 3 1 1
1 1 1 1 1 3 3 1
Explanation:
The values in the given 2D screen indicate colors of the pixels. X and Y are coordinates of the brush, C is the color that should replace the previous color on screen[X][Y] and all surrounding pixels with the same color. Hence all the 2 are replaced with 3.
BFS 方法：想法是使用 BFS 遍历将颜色替换为新颜色。
- 创建一个空队列让我们说Q 。
- 按下输入中给定的像素的起始位置，并对其应用替换颜色。
- 迭代直到Q不为空并弹出前端节点（像素位置）。
- 检查与当前像素相邻的像素，如果有效则将其推入队列（未使用替换颜色进行着色并且与旧颜色具有相同的颜色）。
下面是上述方法的实现：
Python3
# Python3 implementation of the approach
# Function that returns true if
# the given pixel is valid
def isValid(screen, m, n, x, y, prevC, newC):
if x<0 or x>= m\
or y<0 or y>= n or\
screen[x][y]!= prevC\
or screen[x][y]== newC:
return False
return True
# FloodFill function
def floodFill(screen,
m, n, x,
y, prevC, newC):
queue = []
# Append the position of starting
# pixel of the component
queue.append([x, y])
# Color the pixel with the new color
screen[x][y] = newC
# While the queue is not empty i.e. the
# whole component having prevC color
# is not colored with newC color
while queue:
# Dequeue the front node
currPixel = queue.pop()
posX = currPixel[0]
posY = currPixel[1]
# Check if the adjacent
# pixels are valid
if isValid(screen, m, n,
posX + 1, posY,
prevC, newC):
# Color with newC
# if valid and enqueue
screen[posX + 1][posY] = newC
queue.append([posX + 1, posY])
if isValid(screen, m, n,
posX-1, posY,
prevC, newC):
screen[posX-1][posY]= newC
queue.append([posX-1, posY])
if isValid(screen, m, n,
posX, posY + 1,
prevC, newC):
screen[posX][posY + 1]= newC
queue.append([posX, posY + 1])
if isValid(screen, m, n,
posX, posY-1,
prevC, newC):
screen[posX][posY-1]= newC
queue.append([posX, posY-1])
# Driver code
screen =[
[1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 0, 0],
[1, 0, 0, 1, 1, 0, 1, 1],
[1, 2, 2, 2, 2, 0, 1, 0],
[1, 1, 1, 2, 2, 0, 1, 0],
[1, 1, 1, 2, 2, 2, 2, 0],
[1, 1, 1, 1, 1, 2, 1, 1],
[1, 1, 1, 1, 1, 2, 2, 1],
]
# Row of the display
m = len(screen)
# Column of the display
n = len(screen[0])
# Co-ordinate provided by the user
x = 4
y = 4
# Current color at that co-ordinate
prevC = screen[x][y]
# New color that has to be filled
newC = 3
floodFill(screen, m, n, x, y, prevC, newC)
# Printing the updated screen
for i in range(m):
for j in range(n):
print(screen[i][j], end =' ')
print()1 1 1 1 1 1 1 1
1 1 1 1 1 1 0 0
1 0 0 1 1 0 1 1
1 3 3 3 3 0 1 0
1 1 1 3 3 0 1 0
1 1 1 3 3 3 3 0
1 1 1 1 1 3 1 1
1 1 1 1 1 3 3 1DFS 方法：类似的 DFS 方法也可用于实现 Flood Fill 算法。
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。