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).
此题深受Meeting Rooms2影响，用了同样的方法，后来感觉优先队列似乎有些多余，可以进行简化，没简化的代码如下：
public class Solution {
public int findMinArrowShots(int[][] points) {
if(points==null||points.length==0) return 0;
Arrays.sort(points,new Comparator<int[]>(){
public int compare(int[] p1,int[] p2){
return p1[0]-p2[0];
}
});
PriorityQueue<int[]> q = new PriorityQueue<int[]>(points.length,new Comparator<int[]>(){
public int compare(int[] p1,int[] p2){
return p2[1]-p1[1];
}
});
q.offer(points[0]);
for(int i=1;i<points.length;i++){
int[] cur = q.poll();
if(points[i][0]<=cur[1]){
cur[0] =points[i][0];
cur[1] = Math.min(points[i][1],cur[1]);
}else{
q.offer(points[i]);
}
q.offer(cur);
}
return q.size();
}
}后来看了答案，修改了代码如下：
public class Solution {
public int findMinArrowShots(int[][] points) {
if(points==null||points.length==0) return 0;
Arrays.sort(points,new Comparator<int[]>(){
public int compare(int[] p1,int[] p2){
return p1[0]-p2[0];
}
});
int min = points[0][1];
int count=1;
for(int i=1;i<points.length;i++){
if(points[i][0]<=min){
min = Math.min(points[i][1],min);
}else{
min = points[i][1];
count++;
}
}
return count;
}
}