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>& ops) {

if (ops.empty() || ops[0].empty()) return m * n;

priority_queue r, c;

for (auto op : ops) {

r.push(-op[0]);

c.push(-op[1]);

}

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>& ops) {

for (auto op : ops) {

m = min(m, op[0]);

n = min(n, op[1]);

}

return m * n;

}

};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/598

 

类似题目：

Range Addition

 

参考资料：

https://leetcode.com/problems/range-addition-ii/

https://leetcode.com/problems/range-addition-ii/discuss/103595/Java-Solution-find-Min

 

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