leetcode 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).

我估摸着画了个图。

我的思路就是将气球的按照起点从小到大排序。然后从左到右一个个地用箭戳交叉的部位。将end初始化为第一个气球(设为A)的终点。然后看第二个气球(设为B)的起点：
   如果B的起点已经在A的终点后面了，那么说明两者没有交集，得赶紧戳破气球A了。
   如果B的起点在A的终点前面，说明两者有交集，这样分两种情况：
         1. 两者相交但不是包含关系(图1中有类似的)，说明A的终点小于B的终点，那么需要在A的终点处戳破A，此时B也被戳破了。
         2. 两者处于包含关系(如图2所示)，说明A的终点大于B的终点，那么需要在B的终点处戳破B，此时A也被戳破了。

package leetcode;import java.util.Arrays;public class Minimum_Number_of_Arrows_to_Burst_Balloons_452 {public int findMinArrowShots(int[][] points) {if(points==null||points.length==0){return 0;}int arrows=0;Arrays.sort(points,(a,b) -> a[0]==b[0]? a[1]-b[1] : a[0]-b[0]);int end=points[0][1];arrows+=1;for(int i=1;i<points.length;i++){if(points[i][0]>end){arrows+=1;end=points[i][1];}else{end=Math.min(end, points[i][1]);}}return arrows;}public static void main(String[] args) {// TODO Auto-generated method stubMinimum_Number_of_Arrows_to_Burst_Balloons_452 m=new Minimum_Number_of_Arrows_to_Burst_Balloons_452();int[][] points=new int[][]{{10,16},{2,8},{1,6},{7,12}};System.out.println(m.findMinArrowShots(points));}}

大神称之为greedy solution.