> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/598.range_addition_ii.md).

# 598.Range Addition II

**598.Range Addition II**

难度:Easy

> 给定一个初始元素全部为 0，大小为 `m*n` 的矩阵 M 以及在 M 上的一系列更新操作。 操作用二维数组表示，其中的每个操作用一个含有两个正整数 a 和 b 的数组表示，含义是将所有符合 0 <= i < a 以及 0 <= j < b 的元素 M\[i]\[j] 的值都增加 1。 在执行给定的一系列操作后，你需要返回矩阵中含有最大整数的元素个数。

```
示例 1:

输入: 
m = 3, n = 3
operations = [[2,2],[3,3]]
输出: 4
解释: 
初始状态, M = 
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

执行完操作 [2,2] 后, M = 
[[1, 1, 0],
 [1, 1, 0],
 [0, 0, 0]]

执行完操作 [3,3] 后, M = 
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]

M 中最大的整数是 2, 而且 M 中有4个值为2的元素。因此返回 4。
注意:

m 和 n 的范围是 [1,40000]。
a 的范围是 [1,m]，b 的范围是 [1,n]。
操作数目不超过 10000。
```

直接按题意写：

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

> 执行用时 :8ms, 在所有 C++ 提交中击败了99.62%的用户\
> 内存消耗 :11.4MB, 在所有 C++ 提交中击败了94.19%的用户


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