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:
    爬虫header和cookie
    爬虫代理squid
    response对象
    pyspider中内容选择器常用方法汇总
    非阻塞 sleep
    post请求体过大导致ngx.req.get_post_args()取不到参数体的问题
    常用lua代码块
    nginx静态文件缓存的解决方案
    lua-resty-gearman模块
    非在线PDF转图片!!!
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10211634.html
Copyright © 2011-2022 走看看