LeetCode题目：452. Minimum Number of Arrows to Burst Balloons Add to List

最新推荐文章于 2023-12-07 11:13:09 发布

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最新推荐文章于 2023-12-07 11:13:09 发布

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探讨了如何确定最少次数的垂直箭矢射击以击破所有位于二维空间中水平直径已知的圆形气球的问题。介绍了问题背景及具体示例，并提供了一段C++参考代码实现。

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题目原址：点击打开链接

题目描述：

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 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 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).

参考代码：

static bool mysort(pair<int, int>& a, pair<int, int>& b){
        return a.second==b.second?a.first<b.first:a.second<b.second;
}
class Solution {
public:
    int findMinArrowShots(vector<pair<int, int>>& points) {
        sort(points.begin(), points.end(), mysort);
        int num=0;
        int arrow=-1;
        for(int i = 0; i<points.size(); i++){
            if(arrow!=-1 && points[i].first<=arrow){continue;}  
            arrow = points[i].second;
            num++;
        }
        return num;
    }
};