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.
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：
class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
if (ops.empty() || ops[].empty()) return m * n;
priority_queue<int> r, c;
for (auto op : ops) {
r.push(-op[]);
c.push(-op[]);
}
return r.top() * c.top();
}
};
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：
class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
for (auto op : ops) {
m = min(m, op[]);
n = min(n, op[]);
}
return m * n;
}
};
Github 同步地址:
类似题目：
参考资料：
https://leetcode.com/problems/range-addition-ii/
https://leetcode.com/problems/range-addition-ii/discuss/103595/Java-Solution-find-Min