Leetcode-452. Minimum Number of Arrows to Burst Balloons

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

——————————————————————————————

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).

这个题目比较郁闷，想了很久，包括暴力用O(2^n)来穷举，没想到还是不如贪心。。感觉以后要改变思路，贪心和穷举都试一下。 Your runtime beats 32.52% of java submissions.

public class Solution {
    public int findMinArrowShots(int[][] points) {
        if(points==null || points.length==0 || points[0].length==0) return 0;
        Arrays.sort(points, new Comparator() {
            public int compare(int[] a, int[] b) {
                if(a[0]==b[0]) return a[1]-b[1];
                else return a[0]-b[0];
            }
        });

        int minArrows = 1;
        int arrowLimit = points[0][1];
        for(int i=1;i