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  • 【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
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  • 原文地址:https://www.cnblogs.com/seyjs/p/14793638.html
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