leetcode1252

class Solution:
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        matrix = [[0 for _ in range(m)]for _ in range(n)]
        row,column = [0] * n,[0] * m
        for ri in range(len(indices)):
            row[indices[ri][0]] += 1
            column[indices[ri][1]] += 1
        for r in range(len(row)):
            if row[r] > 0:
                for j in range(m):
                    matrix[r][j] += row[r]
        for c in range(len(column)):
            if column[c] > 0:
                for i in range(n):
                    matrix[i][c] += column[c]
        count = 0
        for i in range(n):
            for j in range(m):
                if matrix[i][j] % 2 == 1:
                    count += 1
        return count

这个执行效率偏低，下面进行改进：

class Solution:
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        matrix = [[0 for _ in range(m)]for _ in range(n)]
        row,column = [0] * n,[0] * m
        count = 0
        for ri in range(len(indices)):
            row[indices[ri][0]] += 1
            column[indices[ri][1]] += 1
        for i in range(n):
            for j in range(m):
                if (row[i] + column[j]) % 2 == 1:
                    count += 1
        
        return count

思路：先计算“行”和“列”分别增加了几次。

然后遍历整个二维数组，计算每个元素加1的次数(对应位置的行加1的次数+对应位置列加1的次数)。

增加后，判断此位置是否为奇数(odd)，统计所有奇数的个数存储在count中。最后返回count。