leetcode 452. Minimum Number of Arrows to Burst Balloons 消除覆盖区间

忘是亡心i 2022-06-04 09:07 125阅读 0赞

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).

本题题意很简单，就是寻找可以射击气球的箭的数量，其实和[leetcode 435. Non-overlapping Intervals 消除覆盖区间][leetcode 435. Non-overlapping Intervals] 是一个问题，就是寻找公共区间，然后统计即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    
    bool cmp(pair<int, int> a, pair<int, int> b)
    {
        if (a.first != b.first)
            return a.first < b.first;
        else
            return a.second < b.second;
    }
    
    class Solution 
    {
    public:
        int findMinArrowShots(vector<pair<int, int>>& p) 
        {
            if (p.size() <= 0)
                return 0;
    
            sort(p.begin(),p.end(),cmp);
            int pre = 0;
            int count = 1;
            for (int i = 1; i < p.size(); i++)
            {
                if (p[i].first > p[pre].second)
                {
                    count++;
                    pre = i;
                }
                else
                {
                    if (p[i].second < p[pre].second)
                        pre = i;
                }
            }
            return count;
        }
    };

[leetcode 435. Non-overlapping Intervals]: http://blog.csdn.net/jackzhang_123/article/details/78759839