LeetCode: 598 Range Addition II(easy)

时间:2023-03-08 16:34:33
LeetCode: 598 Range Addition II(easy)

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:

  1. The range of m and n is [1,40000].
  2. The range of a is [1,m], and the range of b is [1,n].
  3. The range of operations size won't exceed 10,000.

代码:

没太看懂题,稍后再做。

参考:http://www.cnblogs.com/grandyang/p/6974232.html

【这道题看起来像是之前那道Range Addition的拓展,但是感觉实际上更简单一些。每次在ops中给定我们一个横纵坐标,将这个子矩形范围内的数字全部自增1,让我们求最大数字的个数。原数组初始化均为0,那么如果ops为空,没有任何操作,那么直接返回m*n即可,我们可以用一个优先队列来保存最大数字矩阵的横纵坐标,我们可以通过举些例子发现,只有最小数字组成的边界中的数字才会被每次更新,所以我们想让最小的数字到队首,更优先队列的排序机制是大的数字在队首,所以我们对其取相反数,这样我们最后取出两个队列的队首数字相乘即为结果,参见代码如下:】

解法一:

LeetCode: 598 Range Addition II(easy)
class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
if (ops.empty() || ops[0].empty()) return m * n;
priority_queue<int> r, c;
for (auto op : ops) {
r.push(-op[0]);
c.push(-op[1]);
}
return r.top() * c.top();
}
};
LeetCode: 598 Range Addition II(easy)

我们可以对空间进行优化,不使用优先队列,而是每次用ops中的值来更新m和n,取其中较小值,这样遍历完成后,m和n就是最大数矩阵的边界了,参见代码如下:

解法二:

LeetCode: 598 Range Addition II(easy)
class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
for (auto op : ops) {
m = min(m, op[0]);
n = min(n, op[1]);
}
return m * n;
}
};
LeetCode: 598 Range Addition II(easy)