Description： 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:

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

題意：給定兩個正整數m,n構造一個所有元素均為0的矩陣M，同時有一個操作矩陣，每行包括兩個元素a和b，對構造的矩陣執行M[i][j] += 1(0<=i<a, 0<=j<b)；要求計算，經過操作矩陣的操作後，矩陣M中最大值元素的數量；

解法：最簡單的辦法自然就是先構造初始矩陣，然後根據給定的操作矩陣計算最後的矩陣，最後再求最大值的個數；但是，我們看到對於操作矩陣中的兩個元素a,b，我們對構造的矩陣操作的元素範圍為[0,a-1][0,b-1]，因此我們只需要求操作矩陣中的所有操作的最小值，那麼這個範圍內的元素每次都會被加1，因此，最後得到的值也是最大的；

Java

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        int row = m;
        int col = n;
        for (int[] op : ops) {
            int a = op[0];
            int b = op[1];
            if (a < row) row = a;
            if (b < col) col = b;
        }
        return row * col;
    }
}