Leetcode题解

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

链接
贪心法，先把气球的区间按从小到大排列好，根据最后穿过的位置与右侧位置相比去较小值来设定新的最后穿过的位置。然后每次穿过res计数加一。

class Solution {public:int findMinArrowShots(vector<pair<int, int>>& points) {        if (points.empty()) return 0;        sort(points.begin(), points.end());        int res = 1, end = points[0].second;        for (int i = 1; i < points.size(); ++i) {            if (points[i].first <= end) {                end = min(end, points[i].second);            } else {                ++res;                end = points[i].second;            }        }        return res;}};