zoukankan      html  css  js  c++  java
  • 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).

    Approach #1: C++. [80ms]

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

      

    Approach #2: C++. [24ms]

    static int x=[](){
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        return 0;
    }();
    
    class Solution {
    public:
        int findMinArrowShots(vector<pair<int, int>> &points) {
            if (points.size() == 0 || points.size() == 1)
                return points.size();
            sort(points.begin(), points.end(), [](auto &p1, auto &p2) {
                return (p1.second <= p2.second);
            });
            int right = INT_MIN, res = 0;
            for (auto x : points) {
                if (x.first <= right) continue;
                right = x.second;
                res++;
            }
            return res;
        }
    };
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    golang中,new和make的区别
    k8s客户端库
    k8s 拉取私有镜像
    kubernetes-client / python
    k8s集群外go客户端示例
    K8s获取NodePort
    KUBERNETES中的服务发现机制与方式
    Rancher容器目录持久化
    rancher k8s 实现pod弹性伸缩
    在Terminal里,使用Shift+Insert来代替鼠标右键来进行粘贴操作
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10211634.html
Copyright © 2011-2022 走看看