【leetcode】1582. Special Positions in a Binary Matrix

题目如下：

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][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[i].length
• 1 <= rows, cols <= 100
• mat[i][j] is 0 or 1.

解题思路：很简单的题目。我的方法是先把各行各列的和算出来，然后遍历matrix，如果matrix[i][j] = 1 并且第i行以及第j列的和都为1，则表示这是一个special position。

代码如下：

class Solution(object):
    def numSpecial(self, mat):
        """
        :type mat: List[List[int]]
        :rtype: int
        """
        res = 0
        dic_row = {}
        dic_col = {}
        for i in range(len(mat)):
            if sum(mat[i]) == 1:
                dic_row[i] = 1

        for j in range(len(mat[0])):
            amount = 0
            for i in range(len(mat)):
                amount += mat[i][j]
            if amount == 1:dic_col[j] = 1

        for i in range(len(mat)):
            for j in range(len(mat[i])):
                if mat[i][j] == 1 and i in dic_row and j in dic_col:
                    res += 1
        return res