【LeetCode】598. Range Addition II


题目描述:

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.

代码实现:

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        int numl=m,numr=n;
        for(int i=0;i<ops.size();i++){
            numl = numl<ops[i][0]?numl:ops[i][0];
            numr = numr<ops[i][1]?numr:ops[i][1];
        }

        return numl*numr;
    }
};


 
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