用于搜索矩阵中连续出现 n 次的最小元素的Python程序
给定一个包含 n 行的矩阵。每行有相同数量的 m 个元素。任务是找到在矩阵中水平、垂直、对角连续出现 n 次的元素。如果有多个这样的元素，则打印最小的元素。如果没有这样的元素
然后打印-1。
例子：
Input : n = 4
mat[5][5] 2 1 3 4 5
3 2 1 3 4
5 3 2 1 4
5 4 3 2 1
5 4 4 3 2
Output : 1
Input : n = 2
mat[4][4] 2 0 5 1
0 5 3 5
8 4 1 1
3 8 5 2
Output : 0执行：
Python3
def check(l,row,col ,n,l2):
# Iterate through the matrix.
# Check for diagonal.
for i in range(row - n+1):
for j in range(col - n+1):
# Store the value in a temporary
# variable.
num = l[i][j]
# Check for the condition only
# if we have more rows and columns
# than n.
if(row-i >= n and col-j >= n):
count = 0
# check if the number is present
# n times.
for k in range(n):
# if number is not present n
# times than break
if(num != l[i+k][j+k]):
break
# increment the count for checking
# the condition follows.
else:
count += 1
# if count is same or greater as that of
# n than the number follows the condition.
if(count == n):
l2.append(num)
else:
break
#check for row condition
for i in range(row):
for j in range(col - (n-1) ):
num = l[i][j]
count = 0
for k in range(n):
if num != l[i][j + k]:
break
else:
count+=1
# if count is same or greater as that
# of n than the number follows the condition.
if(count==n):
l2.append(num)
#check for column condition.
for i in range(row - (n-1)):
for j in range(col):
num = l[i][j]
count = 0
for k in range(n):
if num != l[i+k][j]:
break
else:
count += 1
# if count is same or greater as that of
# n than the number follows the condition.
if(count == n):
l2.append(num)
# It would require a complex code to
# check for anti-diagonal Just rotate
# the matrix and check for diagonal
# condition again.
#matrix rotation by 90 degrees.
for i in range(0, int(row / 2)):
for j in range(i, col - i - 1):
# store current cell in
# num variable
num = l[i][j]
# move values from right to top
l[i][j] = l[j][col - 1 - i]
# move values from bottom to right
l[j][col - 1 - i] = l[row - 1 - i][col - 1 - j]
# move values from left to bottom
l[row - 1 - i][col - 1 - j] = l[row - 1 - j][i]
# assign num to left
l[row - 1 - j][i] = num
# Iterate through the rotated matrix.
for i in range(row - n+1):
for j in range(col - n+1):
# Store the value in a
# temporary variable.
num = l[i][j]
# Check for the condition only if
# we have more rows and columns than n.
if(row-i >= n and col-j >= n):
count = 0
# check if the number is present
# n times.
for k in range(n):
# if number is not present n
# times than break
if(num != l[i+k][j+k]):
break
# increment the count for checking
# the condition follows.
else:
count += 1
# if count is same or greater as that of
# n than the number follows the condition.
if(count == n):
l2.append(num)
else:
break
# check if any element followed the condition.
if(len(l2) == 0):
print(-1)
# print the minimum of all the elements
# which follow the condition.
else:
print(min(l2))
if __name__ == "__main__":
# Create Matrix
l = [[2, 1, 3, 4, 5],
[0, 2, 1, 3, 4],
[5, 0, 2, 1, 4],
[5, 4, 0, 2, 1],
[5, 4, 4, 0, 2]]
# Create a list to store all the elements
# which follow the condition.
l2 = []
check(l,4,4,2,l2)输出 ：
0