Range Addition II(leetcode)

Range Addition II

Range Addition II
题目

解决

题目

leetcode题目

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.

解决

这里我们需要得到每次操作的最小覆盖。因为变化都是从0开始到预定的数值，这样我们只需遍历给定的

ops

中的每组数对

(a, b)

，分别取

a

和

b

中最小的数，其结果为

a * b

。

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