Leetcode 1252. Cells with Odd Values in a Matrix

文章作者：Tyan 博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

**解析：**Version 1，由于是统计奇数的个数，因此不需要每次加1，只需要奇数次设置值为1，偶数次设置值为0，最后统计矩阵的和即可，通过异或^1实现。Version 2非常巧妙，由于每一次操作都是一行或一列，因此用两个数组就可以模拟所有的行和列，最终矩阵的第(i,j)个元素的值为row[i] col[j]，对矩阵的每个值进行统计即可。Version 1需要修改一行或一列，现在只需要修改一个数值即可，大大减少了操作的数量，运行时间明显减少。

• Version 1

代码语言：javascript复制

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        matrix = [[0] * n for _ in range(m)]
        for row, col in indices:
            for i in range(n):
                matrix[row][i] = matrix[row][i] ^ 1
            for i in range(m):
                matrix[i][col] = matrix[i][col] ^ 1
        return sum([sum(row) for row in matrix])

• Version 2

代码语言：javascript复制

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        rows = [0] * m
        cols = [0] * n
        for i, j in indices:
            rows[i] ^= 1
            cols[j] ^= 1

        count = 0
        for i in range(m):
            for j in range(n):
                if rows[i]   cols[j] == 1:
                    count  = 1
        return count

Reference

1. https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/

sdn com row version 统计

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