[leetcode]452. Minimum Number of Arrows to Burst Balloons

[leetcode]452. Minimum Number of Arrows to Burst Balloons

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Analysis

好冷鸭—— [每天刷题并不难0.0]

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.

首先对输入排序，先按照第二个坐标排序，第二个坐标相同的再按照第一个坐标排序。然后每次都吧arrow放在节点的第二个坐标处，然后判断是否能覆盖下一个节点，不能的话就增加箭头，以此类推

Implement

class Solution {
public:
    int findMinArrowShots(vector<pair<int, int>>& points) {
        int res = 0;
        int arrow = INT_MIN;
        int len = points.size();
        sort(points.begin(), points.end(), mycmp);
        for(int i=0; i<len; i++){
            if(arrow != INT_MIN && points[i].first <= arrow)
                continue;
            arrow = points[i].second;
            res++;
        }
        return res;
    }
    static bool mycmp(pair<int, int>& p1, pair<int, int>& p2){
        if(p1.second == p2.second)
            return p1.first < p2.first;
        else
            return p1.second < p2.second;
    }
};