【leetcode】1252. Cells with Odd Values in a Matrix

题目以下：

Given n and m which are the dimensions of a matrix initialized by zeros and given an array indices where indices[i] = [ri, ci]. For each pair of [ri, ci] you have to increment all cells in row ri and column ci by 1.

Return the number of cells with odd values in the matrix after applying the increment to all indices.

Example 1:

Input: n = 2, m = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.

Example 2:

Input: n = 2, m = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.

Constraints:

• 1 <= n <= 50
• 1 <= m <= 50
• 1 <= indices.length <= 100
• 0 <= indices[i][0] < n
• 0 <= indices[i][1] < m

解题思路：首先遍历indices，计算出每行每列分别作了几回+1的操做。而后再遍历matrix，根据matrix[i][j]求得对应的i行和j列分别作了几回+1，判断二者之和的奇偶性便可。

代码以下：

class Solution(object):
    def oddCells(self, n, m, indices):
        """
        :type n: int
        :type m: int
        :type indices: List[List[int]]
        :rtype: int
        """
        dic_row = {}
        dic_column = {}
        for (r,c) in indices:
            dic_row[r] = dic_row.setdefault(r,0) + 1
            dic_column[c] = dic_column.setdefault(c, 0) + 1
        res = 0
        for i in range(n):
            for j in range(m):
                count = dic_row.get(i,0) + dic_column.get(j,0)
                if count % 2 == 1:res += 1
        return res