There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it‘s horizontal, y-coordinates don‘t matter and hence the x-coordinates of start and end of the diameter suffice. Start is always smaller than end. There will be at most 104 balloons.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.

Example:

Input:
[[10,16], [2,8], [1,6], [7,12]]

Output:
2

Explanation:
One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).

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這個題是貪心題，是區間選點問題。在排序C++的vector時，需要在外面用bool，並且要加上inline這個內聯函數。還需要註意points是一個空數組的可能情況。

C++代碼：

inline bool cmp(const vector<int> &a,const vector<int> &b){
    return a[1] < b[1];
}
class Solution {
public:
    
    int findMinArrowShots(vector<vector<int>>& points) {
        if(points.size() == 0)
            return 0;
        sort(points.begin(),points.end(),cmp);
        int sum = 1;
        int ans = points[0][1];
        for(int i = 1; i < points.size(); i++){
            if(ans < points[i][0]){
                sum++;
                ans = points[i][1];
            }
        }
        return sum;
    }
};

(貪心) leetcode 452. Minimum Number of Arrows to Burst Balloons