Leetcode 452. Minimum Number of Arrows to Burst Balloons

题目：

here 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:2Explanation:One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at

思路：

求解不想交的区域。因为不想交的区域至少需要一只箭才能消灭。如果遇见在后面一个区域与前一个相交，判断是否该区域被前面一个包含，如果包含就把它作为当前区域作，没有就向后遍历，直到不想交时，count++；遍历到结束就行了。

bool cmp(pair<int,int> a,pair<int,int> b) {    return a.first<b.first?1:0;}class Solution {public:        int findMinArrowShots(vector<pair<int, int>>& points) {        if(points.size() == 0) return 0;        sort(points.begin(),points.end(),cmp);        int i = 1;        int y = points[0].second;        int count = 1;        while(i<points.size()){            if(points[i].first<=y) {                if(points[i].second<=y) {                    y = points[i].second;                }                i++;            } else {                count ++;                y = points[i].second;                i++;            }        }        return count;    }};