算法分析与设计week06--452.Minimum Number of Arrows to Burst Balloons

452.Minimum Number of Arrows to Burst Balloons

Description

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).

Analyse

思路：先对开始坐标进行排序。当遇到一个结束坐标时，该气球必须被戳爆，顺带戳爆其他的气球。

相邻两个气球之间有三种情况：
1.没有重叠
2.有重叠，但后者不完全包含在前者中
3.后者完全包含在前者中

1.如果两者没有重叠，则需要再加一枝箭；
2.如果两者有重叠且后者不完全包含在前者中，在重叠部分射箭可以节省一支箭，拿重叠部分继续和后续气球做比对；
3.如果后者完全包含在前者中，在后者区间内射箭可以节省一支箭，拿后者继续和后续气球做比对。

通过以上策略可以使箭的数量达到最少。后两种情况可以合并为一种情况考虑，这里为了更容易理解分为两类来处理。

class Solution {public:    int findMinArrowShots(vector<pair<int, int>>& points) {        if (points.size() == 0) {            return 0;        }        //  排序        sort(points.begin(), points.end());        int arrow = 1;        int end = points[0].second;        for (int i = 1; i < points.size(); i++) {            //  第一种情况            if(points[i].first > end) {                arrow++;                end = points[i].second;            }            //  第二、三种情况            else {                end = min(end, points[i].second);            }        }         return arrow;    }};

复杂度分析

时间复杂度：O(nlogn)
空间复杂度：O(1)