Minimum Number of Arrows to Burst Balloons

最新推荐文章于 2023-11-18 01:25:37 发布

小灰灰52309

最新推荐文章于 2023-11-18 01:25:37 发布

阅读量235

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode ACM 文章标签： Minimum Number of Ar

本文链接：https://blog.youkuaiyun.com/xiaohuihui52309/article/details/70316911

leetcode 同时被 2 个专栏收录

185 篇文章

订阅专栏

ACM

127 篇文章

订阅专栏

探讨了在二维空间中，如何确定射箭的最小数量以击破所有气球的问题。通过将气球的水平直径坐标排序并采用特定算法，实现了用最少的箭矢击破所有目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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).

解析：仔细分析最少使用多少箭头能打破所有气球，对所有气球按start位置进行排序，然后从后向前遍历每个气球，每次把气球插在起始点位置上，只要前一个气球的end大小箭头位置，则不需要重新插气球，否则需要再次插个气球。

代码：

class Solution {
public:

     static  bool cmp(pair<int,int>p1,pair<int,int>p2)
    {
        return p1.first<p2.first?1:(p1.first==p2.first?p1.second<p2.second:0);
    }


    int findMinArrowShots(vector<pair<int, int>>& points) {
        
        int ans=0;
        int arrow=INT_MIN;
        if (points.empty()) return ans;
        sort(points.begin(),points.end(),cmp);
        ans=1;
        arrow=points[points.size()-1].first;
        for (int i=points.size()-2; i>=0; i--)
        {
            if (points[i].second>=arrow) continue;
            ans++;
            arrow=points[i].first;
        }
        return ans;
    }
};