C++ 贪心算法射击气球

已知一个平面上有一定数量的气球，平面可以看做一个坐标系，在平面的x轴的不同位置安排弓箭手向y轴方向射箭，弓箭可以向y轴走无穷远；给定气球的宽度xstart<=x<=xend。问至少需要多少弓箭手，将全部气球打爆？
例如：四个气球: [[10,16],[2,8],[1,6],[7,12]]，至少需要2个弓箭手。

#include<algorithm>
#include<vector>
bool cmp(const std::pair<int, int>& a, const std::pair<int, int>& b)
{
 return a.first < b.first;
}
class Solution 
{
public:
 int findMinArrowShots(std::vector<std::pair<int, int>>& points)
 {
  if (points.size()==0)
  {
   return 0;
  }
  std::sort(points.begin(), points.end(), cmp);
  int shoot_num = 1;
  int shoot_begin = points[0].first;
  int shoot_end = points[0].second;
  for (unsigned i = 0; i < points.size(); i++)
  {
   if (points[i].first <= shoot_end)
   {
    shoot_begin = points[i].first;
    if (shoot_end>points[i].second)
    {
     shoot_end = points[i].second;
    }
   }
   else
   {
    shoot_num++;
    shoot_begin = points[i].first;
    shoot_end = points[i].second;
   }
  }
  return shoot_num;
 }
};
int main()
{
 std::vector<std::pair<int, int>> points;
 points.push_back(std::make_pair(10, 16));
 points.push_back(std::make_pair(2, 8));
 points.push_back(std::make_pair(1, 6));
 points.push_back(std::make_pair(7, 12));
 Solution solve;
 printf("%d\n", solve.findMinArrowShots(points));
 return 0;
}

运行结果：
2