[LeetCode] Range Addition II

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.

一个m*n的全0矩阵在两个变换后的最大值的数目。每次变换是在一定范围内矩阵中的值+1。如果用每一步的操作来解决，那么提交时会显示超时。那么用一个巧妙的方法来解决。每一次操作都会涉及矩阵左上角的一个矩形，只需要计算两次操作矩形的重叠面积即可。这个矩形的长宽是有两次操作的长宽与原矩阵的长宽的最小值决定。

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;
    }
};
// 6 ms

 

转载于:https://www.cnblogs.com/immjc/p/7161058.html