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 <= 501 <= m <= 501 <= indices.length <= 1000 <= indices[i][0] < n0 <= indices[i][1] < mclass Solution { public int oddCells(int n, int m, int[][] indices) { int[][] res = new int[n][m]; for(int i = 0; i < n; i++) Arrays.fill(res[i], 0); for(int i = 0; i < indices.length; i++){ int r = indices[i][0]; int c = indices[i][1]; for(int j = 0; j < m; j++) res[r][j]++; for(int j = 0; j < n; j++) res[j][c]++; } int re = 0; for(int i = 0; i < n; i++){ for(int j = 0; j < m; j++){ if(res[i][j] % 2 != 0) re++; } } return re; } }
题意思是初始给一个全0数组，然后给一个indices数组里面的ri，ci表示row和column，然后在该row，column每个increment1.最后算数组有几个odd
另一种方法
class Solution { public int oddCells(int n, int m, int[][] indices) { if(n <= 0 || m <= 0 || indices == null || indices.length == 0 || indices[0].length == 0) return 0; int[] rows = new int[n]; int[] clms = new int[m]; for(int i = 0; i < indices.length; i++){ int[] rc = indices[i]; rows[rc[0]]++; clms[rc[1]]++; } int ret = 0; for(int r = 0; r < rows.length; r++){ for(int c = 0; c < clms.length; c++){ if((rows[r] + clms[c]) % 2 == 1){ ret++; } } } return ret; } }
一维数组比二维数组快，学到了。
1252. Cells with Odd Values in a Matrix
原文：https://www.cnblogs.com/wentiliangkaihua/p/11870425.html