leetcode 598. Range Addition II

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.

本题题意看起来十分的复杂，其实十分的简单，只需要遍历ops的值来更新m和n，求最小值，这个就是最终的最大数的边界

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>using namespace std;class Solution {public:    int maxCount(int m, int n, vector<vector<int>>& ops)    {        for (vector<int> a :ops)        {            m = min(m, a[0]);            n = min(n, a[1]);        }        return m*n;    }};