LeetCode 598. Range Addition II

598. Range Addition II

Descripition

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:Input: m = 3, n = 3operations = [[2,2],[3,3]]Output: 4Explanation: 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.

Solution

• 题意即给一个初始化的数组（均为零），以及一些二元组，要求将二元组范围内的（见题目描述）数加一。最终返回最大数的数目。
• 分析题目，我们可以发现，不论二元组是什么，总是从（0，0）开始累加，这就启发我们去找这些二元组的交集。交集内的数一定是最大数，我们可以获得二元组各个元素的最小值，结果便是这些最小值的乘积，代码如下。

class Solution {public:    long long maxCount(int m, int n, vector<vector<int>>& ops) {        if (ops.size() == 0) return m * n;    //如果未给出二元组，那么返回原始数组的大小，因为0就是最大值。        int min_row = ops[0][0],min_col = ops[0][1];        for (int i = 1;i < ops.size();i++) {            min_row = min(min_row,ops[i][0]);            min_col = min(min_col,ops[i][1]);        }        return min_row * min_col;    }};