|
马上注册,结交更多好友,享用更多功能^_^
您需要 登录 才可以下载或查看,没有账号?立即注册
x
Given a rows x cols matrix mat, where mat[j] is either 0 or 1, return the number of special positions in mat.
A position (i,j) is called special if mat[j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).
Example 1:
Input: mat = [[1,0,0],
[0,0,1],
[1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.
Example 2:
Input: mat = [[1,0,0],
[0,1,0],
[0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions.
Example 3:
Input: mat = [[0,0,0,1],
[1,0,0,0],
[0,1,1,0],
[0,0,0,0]]
Output: 2
Example 4:
Input: mat = [[0,0,0,0,0],
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,1,0,0],
[0,0,0,1,1]]
Output: 3
Constraints:
rows == mat.length
cols == mat.length
1 <= rows, cols <= 100
mat[j] is 0 or 1.
- class Solution:
- def numSpecial(self, mat: List[List[int]]) -> int:
- m = len(mat)
- n = len(mat[0])
- res = 0
-
- for i in range(m):
- for j in range(n):
- if mat[i][j] == 1:
- count1 = 0
- count2 = 0
- for k in range(n):
- if mat[i][k] == 0:
- count1 += 1
-
- for k in range(m):
- if mat[k][j] == 0:
- count2 += 1
-
- if count1 == n - 1 and count2 == m - 1:
- res += 1
- return res
复制代码 |
|