【leetcode】598. Range Addition II

　　You are given an m x n matrix M initialized with all 0's and an array of operations ops, where ops[i] = [ai, bi] means M[x][y] should be incremented by one for all 0 <= x < ai and 0 <= y < bi.

Count and return the number of maximum integers in the matrix after performing all the operations.

class Solution {

public:

    int maxCount(int m, int n, vector<vector<int>>& ops) {

        //取所有操作的交集就是最大int number的个数 冷静思考就没问题 啊哈哈哈

        if(ops.size()==0) return m*n;

        int limit_m=m,limit_n=n;

        for(auto aa:ops)

        {

            limit_m=min(limit_m,aa[0]);

            limit_n=min(limit_n,aa[1]);

        }

        return limit_m*limit_n;

    }

};

巴特西

【leetcode】598. Range Addition II

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