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# [学习笔记] Leetcode 1582. Special Positions in a Binary Matrix 发表于 2020-9-16 21:20:22 | 显示全部楼层 |阅读模式

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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 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.

1. class Solution:
2.     def numSpecial(self, mat: List[List[int]]) -> int:
3.         m = len(mat)
4.         n = len(mat)
5.         res = 0
6.
7.         for i in range(m):
8.             for j in range(n):
9.                 if mat[i][j] == 1:
10.                     count1 = 0
11.                     count2 = 0
12.                     for k in range(n):
13.                         if mat[i][k] == 0:
14.                             count1 += 1
15.
16.                     for k in range(m):
17.                         if mat[k][j] == 0:
18.                             count2 += 1
19.
20.                     if count1 == n - 1 and count2 == m - 1:
21.                         res += 1
22.         return res

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