Range Addition II(leetcode)

Range Addition II

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题目

leetcode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 = 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.

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.

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解决

这里我们需要得到每次操作的最小覆盖。因为变化都是从0开始到预定的数值，这样我们只需遍历给定的ops中的每组数对(a, b)，分别取a和b中最小的数，其结果为a * b。

class Solution {public:    int maxCount(int m, int n, vector<vector<int>>& ops) {        int a = m;        int b = n;        for (int i = 0; i < ops.size(); i++) {            a = min(a, ops[i][0]);            b = min(b, ops[i][1]);        }        return a * b;    }};