452. Minimum Number of Arrows to Burst Balloons

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: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 x = 11 (bursting the other two balloons).

贪心法则，这题本质上和题目 435. Non-overlapping Intervals 是一样的

对于有部分重叠的区间（线段），每次只保留一个，留下一个右端点“最靠右”的区间（线段），对删除的区间（线段）进行计数，总共删除了K个区间，最后的答案等于points.length()-K

public class Solution {    public static int findMinArrowShots(int[][] points){int m=points.length;if(m<1)return 0;Arrays.sort(points,new Comparator<int[]>(){@Overridepublic int compare(int[] o1, int[] o2){// TODO Auto-generated method stubif(o1[0]!=o2[0])return o1[0]-o2[0];return o1[1]-o2[1];}});int ret=0;int prev=points[0][1];for(int i=1;i<m;i++){if(points[i][0]<=prev){ret++;if(prev<=points[i][1])continue;}prev=points[i][1];}return m-ret;}}