598. Range Addition II

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本文介绍了一个LeetCode上的算法题目，该题目要求计算一系列更新操作后矩阵中最大整数的数量。通过寻找所有操作中的最小行数和列数来确定最大整数所在的子矩阵范围。

题目来源【Leetcode】

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.

这道题就是找到操作里面最小行和列，他们组成的就死最大的数的矩阵，代码如下：

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        int mr = m;
        int mc = n;
        for(int i = 0; i < ops.size(); i++){
            mr = min(ops[i][0],mr);
            mc = min(ops[i][1],mc);
        }
        return mr*mc;
    }
};