题目来源
这道题感觉不难，先排个序，然后从左向右扫描，保存扫描过的气球最大左边界和最小右边界，当最大左边界小于最小右边界时，需要射一箭。

class Solution {
public:
    int findMinArrowShots(vector<pair<int, int>>& points) {
        sort(points.begin(), points.end());
        int n = points.size(), res = 0;
        for (int i=0; i<n; i++) {
            int left_max = points[i].first, right_min = points[i].second;
            while (left_max <= right_min) {
                i++;
                left_max = max(left_max, points[i].first);
                right_min = min(right_min, points[i].second);
            }
            i--;
            res++;
        }
        return res;
    }
};