598. Range Addition II的C++解法
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因为是从(0,0)开始更新,所以只要找到最小的横纵坐标就可以了。注意处理ops为空的情况。
class Solution {public:int maxCount(int m, int n, vector<vector<int>>& ops) {int min1 = m;int min2 = n;for (int i = 0; i < ops.size(); i++){if (ops[i][0] < min1) min1 = ops[i][0];if (ops[i][1] < min2) min2 = ops[i][1];}return min1*min2;}};
题解里的另一种写法:
class Solution {public: int maxCount(int m, int n, vector<vector<int>>& ops) { for (auto op : ops) { m = min(op[0], m); n = min(op[1], n); } return m * n; }};
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