The force deduction method summarizes the number of 1252 odd value cells

  Directory links ：

Force buckle programming problem - The solution sums up _ Share + Record -CSDN Blog

GitHub Synchronous question brushing items ：

https://github.com/September26/java-algorithms

Original link ： Power button

describe ：

To give you one m x n Matrix , At the very beginning , The value in each cell is 0.

There is another two-dimensional index array  indices,indices[i] = [ri, ci] Point to a position in the matrix , among ri and ci Respectively represent the specified row and column （ from 0 Numbered starting ）.

Yes indices[i] Every position pointed , The following incremental operations should be performed at the same time ：

ri All cells on the row , Add 1 .
ci All cells on the column , Add 1 .
Here you are. m、n and indices . Please finish executing all  indices  After the specified incremental operation , Back in the matrix Odd cells Number of .

Example 1：

Input ：m = 2, n = 3, indices = [[0,1],[1,1]]
Output ：6
explain ： The initial matrix is [[0,0,0],[0,0,0]].
After the first incremental operation [[1,2,1],[0,1,0]].
The final matrix is [[1,3,1],[1,3,1]], There are 6 An odd number .
Example 2：

Input ：m = 2, n = 2, indices = [[1,1],[0,0]]
Output ：0
explain ： The final matrix is [[2,2],[2,2]], There are no odd numbers in it .
 

Tips ：

1 <= m, n <= 50
1 <= indices.length <= 100
0 <= ri < m
0 <= ci < n
 

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/cells-with-odd-values-in-a-matrix
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Their thinking ：

*  Their thinking ：
*  This question is relatively simple , Find the last array according to the rules , Then calculate the odd number .

Code ：

public class Solution1252 {

    public int oddCells(int m, int n, int[][] indices) {
        int[][] intss = new int[m][n];

        for (int[] indice : indices) {
            int i1 = indice[0];
            int i2 = indice[1];
            for (int i = 0; i < m; i++) {
                intss[i][i2]++;
            }
            for (int i = 0; i < n; i++) {
                intss[i1][i]++;
            }
        }
        int result = 0;
        for (int i = 0; i < intss.length; i++) {
            for (int k = 0; k < intss[0].length; k++) {
                if (intss[i][k] % 2 != 0) {
                    result++;
                }
            }
        }
        return result;
    }
}