[leetcode]第12周作业

问题：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).

分析

这道题跟贪心算法的典型例题：活动安排，是一样的道理，先把活动按照结束时间排序，选取第一个，然后依次排除与前一个不相容的活动，最后即可得到最多容纳的活动数。在这里，射箭次数要最少，说明每次射中的个数要多，要在尽量多的范围段内。

代码

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