Leetcode: 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:
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).

我的Greedy+Heap做法：

Array按start moment升序sort，Heap里按end moment升序sort，一旦到Heap里结束时间最早的气球的end，就要扔arrow，这时在heap里的气球都会被引爆

 public class Solution {
     public int findMinArrowShots(int[][] points) {
         if (points==null || points.length==0 || points[0].length==0) return 0;
         if (points[0][0]==Integer.MIN_VALUE && points[0][1]==Integer.MAX_VALUE) return 1;
         int res = 0;
         Arrays.sort(points, new Comparator<int[]>() {
             public int compare(int[] a1, int[] a2) {
                 return a1[0]!=a2[0]? a1[0]-a2[0] : a1[1]-a2[1];
             }
         });
         PriorityQueue<int[]> queue = new PriorityQueue<int[]>(1, new Comparator<int[]>() {
             public int compare(int[] a1, int[] a2) {
                 return a1[1]-a2[1];
             }
         });

         for (int[] point : points) {
             if (queue.isEmpty() || point[0] <= queue.peek()[1]) {
                 queue.offer(point);
             }
             else if (point[0] > queue.peek()[1]) {
                 res++;
                 queue.clear();
                 queue.offer(point);
             }
         }
         if (!queue.isEmpty()) res++;
         return res;
     }
 }

改进：因为一旦气球爆了的话，Heap里面的元素要全部清空，不需要知道Heap里面次小值是什么，所以其实不需要Heap, 维护一个最小值即可

 public int findMinArrowShots(int[][] points) {
     if(points==null || points.length==0 || points[0].length==0) return 0;
     Arrays.sort(points, new Comparator<int[]>() {
         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<points.length;i++) {
         int[] baloon = points[i];
         if(baloon[0]<=arrowLimit) {
             arrowLimit=Math.min(arrowLimit, baloon[1]);
         } else {
             minArrows++;
             arrowLimit=baloon[1];
         }
     }
     return minArrows;
 }