題目：

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 positiveintegers 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.

程式碼：

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        int m1 = Integer.MAX_VALUE;
        int m2 = Integer.MAX_VALUE;
        for(int i = 0; i < ops.length; i++){
            if(ops[i][0] < m1) m1 = ops[i][0];
            if(ops[i][1] < m2) m2 = ops[i][1];
        }
        if(m1 > m) m1 = m;
        if(m2 > n) m2 = n;
        return m1 * m2;
    }
}