Given an m * n matrix M initialized with all 0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which meansM[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: m = 3, n = 3operations = [[2,2],[3,3]]Output: 4Explanation: Initially, M = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]After performing [2,2], M = [[1, 1, 0], [1, 1, 0], [0, 0, 0]]After performing [3,3], M = [[2, 2, 1], [2, 2, 1], [1, 1, 1]]So the maximum integer in M is 2, and there are four of it in M. So return 4.

Note:

1. The range of m and n is [1,40000].
2. The range of a is [1,m], and the range of b is [1,n].
3. The range of operations size won't exceed 10,000.

题意：给定一个m*n的矩阵，把数组初始化为0，然后用ops二维数组去更新矩阵的每个元素值，例如a=2,b=2则矩阵M

11

-M

22

全部增加1，以此类推，然后找出最大的元素的个数。

解法：寻找到被操作次数最多的行和列

1. public class Solution {
2. public int MaxCount(int m, int n, int[,] ops) {
3. int rowMin = Int32.MaxValue;
4. int colMin = Int32.MaxValue;
5. if (ops.GetLength(0) == 0) {
6. return m * n;
7. }
8. for (int i = 0; i < ops.GetLength(0); i++) {
9. int row = ops[i, 0];
10. int col = ops[i, 1];
11. if (row > 0) {
12. rowMin = Math.Min(rowMin, row);
13. } else {
14. return 0;
15. }
16. if (col > 0) {
17. colMin = Math.Min(colMin, col);
18. }else {
19. return 0;
20. }
21. }
22. return rowMin * colMin;
23. }
24. }