598. Range Addition II

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Given an m * n matrixMinitialized with all0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with twopositiveintegersaandb, which meansM[i][j]should beadded by onefor all0 <= i < aand0 <= 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.

给定一个M*N 的矩阵和一组操作符operations，每一次将[0,0]—>operation[a,b]的数值加1，求操作后矩阵最大整数的数量。

这不就是求操作符operations中，第一维度和第二维度最小值的乘积么？a*b

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        if (ops == null || ops.length == 0) {
            return m * n;
        }
        int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE;
        for(int op[] : ops)
        {
            row = Math.min(row,op[0]);
            col = Math.min(col,op[1]);
        }
        return row * col;
    }
}