leetcode452. Minimum Number of Arrows to Burst Balloons

题面

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:2Explanation: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).

分析

求需要射箭的最小次数，可以考虑通过贪心算法加以解决。
将所有的气球所在的区间按照起始位置升序排序，想一下，要想射箭的次数最少，那么我们每一次射的箭需要打破尽可能多的气球。而要想打破尽可能多的气球，无非就是选择射箭的位置。
气球的分布已经按照开始位置升序排序，那么，可以在遍历这个序列的同时维护一个begin和end变量来限制射箭的位置，只要箭射在这个区间中，就可以将之前的气球全部打破，然后以下一个气球所在位置为起点，继续遍历，直到数组完毕。

代码

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

运行结果