598. Range Addition II 范围相加

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 = 3
operations = [[2,2],[3,3]]
Output: 4
Explanation: 
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M11-M22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. }







原文地址:https://www.cnblogs.com/xiejunzhao/p/2ef28b4838e32b29f4a8886245304564.html