LeetCode | 1380. Lucky Numbers in a Matrix矩阵中的幸运数【Python】
LeetCode 1380. Lucky Numbers in a Matrix矩阵中的幸运数【Easy】【Python】【暴力】
Problem
Given a m * n
matrix of distinct numbers, return all lucky numbers in the matrix in any order.
A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.
Example 1:
Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
Output: [15]
Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column
Example 2:
Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
Output: [12]
Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.
Example 3:
Input: matrix = [[7,8],[1,2]]
Output: [7]
Constraints:
m == mat.length
n == mat[i].length
1 <= n, m <= 50
1 <= matrix[i][j] <= 10^5
.- All elements in the matrix are distinct.
问题
给你一个 m * n 的矩阵,矩阵中的数字 各不相同 。请你按 任意 顺序返回矩阵中的所有幸运数。
幸运数是指矩阵中满足同时下列两个条件的元素:
- 在同一行的所有元素中最小
- 在同一列的所有元素中最大
示例 1:
输入:matrix = [[3,7,8],[9,11,13],[15,16,17]]
输出:[15]
解释:15 是唯一的幸运数,因为它是其所在行中的最小值,也是所在列中的最大值。
示例 2:
输入:matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
输出:[12]
解释:12 是唯一的幸运数,因为它是其所在行中的最小值,也是所在列中的最大值。
示例 3:
输入:matrix = [[7,8],[1,2]]
输出:[7]
提示:
m == mat.length
n == mat[i].length
1 <= n, m <= 50
1 <= matrix[i][j] <= 10^5
- 矩阵中的所有元素都是不同的
思路
暴力
解法一
找出每一行的最小值,再判断是否是当前列的最大值。
时间复杂度: O(m*n),m 是 matrix 的行数,n 是 matrix 的列数。
Python3代码
from typing import List
class Solution:
def luckyNumbers (self, matrix: List[List[int]]) -> List[int]:
# solution one
m, n = len(matrix), len(matrix[0])
flag = 0
res = []
for i in range(m):
min_ = max_ = matrix[i][0]
for j in range(n):
if matrix[i][j] < min_:
flag = j
min_ = matrix[i][j]
max_ = min_
for x in range(m):
if matrix[x][flag] > max_:
break
elif x == m - 1:
res.append(max_)
return res
解法二
分别找出每一行的最小值和每一列的最大值,再判断是否相等。
时间复杂度: O(max(m, n)),m 是 matrix 的行数,n 是 matrix 的列数。
Python3代码
from typing import List
class Solution:
def luckyNumbers (self, matrix: List[List[int]]) -> List[int]:
# solution two
min_ = {min(rows) for rows in matrix}
max_ = {max(columns) for columns in zip(*matrix)} # zip(*) 对矩阵进行转置,即找出每一列中的最大值
return list(min_ & max_)
代码地址
最怕一生碌碌无为,还说平凡难能可贵。