There are some 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. The start is always smaller than the end.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with x_start and x_end bursts by an arrow shot at x if x_start ≤ x ≤ x_end. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.

Given an array points where points[i] = [x_start, x_end], return the minimum number of arrows that must be shot to burst all balloons.

Example 1:

Input: points = [[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).

Example 2:

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4

Example 3:

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2

Example 4:

Input: points = [[1,2]]
Output: 1

Example 5:

Input: points = [[2,3],[2,3]]
Output: 1

Constraints:

0 <= points.length <= 10^4
points.length == 2
-2^31 <= x_start < x_end <= 2^31 - 1

Solution

Time complexity : O(nlog(n))
Space complexity : O(1)

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

將所有的氣球，依據結束位置由小至大排序。
再每結束位置之內，都射一發箭，即可。