## 题目描述

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 means `M[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.

## 解法一

### 思路 ### Python

``````class Solution:
def maxCount(self, m, n, ops):
"""
:type m: int
:type n: int
:type ops: List[List[int]]
:rtype: int
"""
result_m = m
result_n = n
for op in ops:
result_m = min(result_m, op)
result_n = min(result_n, op)
return result_m * result_n``````

### Java

``````public class Solution {
public int maxCount(int m, int n, int[][] ops) {
int resultM = m;
int resultN = n;
for (int[] op : ops) {
resultM = Math.min(resultM, op);
resultN = Math.min(resultN, op);
}
return resultM * resultN;
}
}``````

### C++

``````class Solution {
public:
int maxCount(int m, int n, vector<vector<int>> &ops) {
int resultM = m;
int resultN = n;
for (const vector<int> &op : ops) {
resultM = min(resultM, op);
resultN = min(resultN, op);
}
return resultM * resultN;
}
};``````