LeetCode 452. Minimum Number of Arrows to Burst Balloons

Posted 2021-03-22 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/

题目：

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).

题解：

Try to get as many ballons as possible with one shot.

First sort the ballons based on ending position. We shot as right as possible with the first ballon, then all following ballons starting <= first ending position are also bursted.

Update the maximum position and skip the following ballons starting <= maximum position.

If current ballon starting position > maximum position, that is the ballon previous arrow can\'t touch, res++, update maximum position.

Note: Sort on the end. Be careful about overfollow.

 

Time Complexity: O(nlogn). n = points.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int findMinArrowShots(int[][] points) {
 3         if(points == null){
 4             return 0;
 5         }
 6         
 7         Arrays.sort(points, (a, b) -> a[1] - b[1]);
 8         int res = 0;
 9         long max = Long.MIN_VALUE;
10         
11         for(int [] arr : points){
12             if(arr[0] <= max){
13                 continue;
14             }
15             
16             res++;
17             max = arr[1];
18         }
19         
20         return res;
21     }
22 }

类似Meeting Rooms II, Non-overlapping Intervals.