452. Minimum Number of Arrows to Burst Balloons

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).

这道题先对二位置的数排序，排序怎么改变sort原有的排序方式见代码，然后就比较简单了，当下一个pair的第一个数<当前pair的第二个数的时候，他们可以用一把剑，而其他情况，换一把剑，代码如下：

class Solution {public:    // bool cmp(pair<int,int>points1,pair<int,int>points2){    //     return points1.second<points2.second;    // }    int findMinArrowShots(vector<pair<int, int>>& points) {        if(points.size()==0)return 0;        sort(points.begin(),points.end(),[](pair<int,int>a,pair<int,int> b){return a.second<b.second;});//哇，这里排序函数的重定义要学学        int i=0;        int cnt=1;        int mark=i;        while(i<points.size()){            if(points[mark].second<points[i].first){                cnt++;                mark=i;            }            i++;        }        return cnt;    }};