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.
题目大意：给出矩阵大小，给出每次加的坐标(x,y)那么每次操作都会使(0~x,0~y)这个矩阵内的格子加1，求最大的格子的个数。
class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
vector<pair<int, int> > v;
if (ops.size() == 0) return m*n;
int x = 100000;
int y = 100000;
for (int i = 0; i < ops.size(); ++i) {
x = min(x, ops[i][0]);
y = min(y, ops[i][1]);
}
return x*y;
}
};