452. Minimum Number of Arrows to Burst Balloons

问题：

给定一组[start, end]表示气球左右边界的气球数组，求最少射多少箭，能将所有气球射爆。

射箭位置X，start<=X<=end，则能射破该气球。

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).

　　

解法：Greedy

同435. Non-overlapping Intervals  

本问题，要求最少射箭次数，使得覆盖全部气球。

假设互相不重叠的气球共有n个，那么最少射n次。

也就将该问题转换为，求有多少个不重叠的气球。

即最多有多少个不重叠区间。

1.按照end排序sort气球。

2.尽量选择end先结束的作为一个不重叠气球。

3.再用上一个不重叠气球的end，来找下一个start>end的气球，作为下一个不重叠气球。

4.最终找完最多不重叠的气球个数。

代码参考：

class Solution {
public:
    bool static cmp(vector<int> a, vector<int> b) {
        return a[1]<b[1];
    }
    //to find the minimal number of arrow shots
    //means one shots to get max number of ballons which are all overlapping.
    //but if there are mostly n non-overlapping ballons, 
    //however we choose the shot position, we should at least shot n times
    // to ensure these n non-overlapping ballons should be all burst.
    //then this problem == 435
    int findMinArrowShots(vector<vector<int>>& points) {
        int count = 1;
        if(points.empty()) return 0;
        sort(points.begin(), points.end(), cmp);
        int pre_end = points[0][1];
        for(int i=1; i<points.size(); i++) {
            if(points[i][0]>pre_end) {
                count++;
                pre_end = points[i][1];
            }
        }
        return count;
    }
};