[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:
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).

解法： 把“气球”看成是一个线段，实际上一个“箭”的坐标就是一个点，要尽可能的被多个线段覆盖到。对于线段先进行排序，然后遍历一遍。注意遍历的时候，要要更新边界值，当下一个线段的起始值大于边界值的时候，表示要新增一个“箭”。

public class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points.length == 0){
            return 0;
        }else if (points.length == 1) {
            return 1;
        } else {
            Arrays.sort(points, (o1, o2) -> o1[0] - o2[0]);
            int count = 1;
            int upperBound = points[0][1];
            for (int i = 0; i < points.length - 1; ) {
                while ( i < points.length && points[i][0]<= upperBound ){
                    if (points[i][1] < upperBound){
                        upperBound = points[i][1];
                    }
                    i++;
                }
                if (i == points.length){
                    break;
                }else {
                    count++;
                    upperBound = points[i][1];
                }
            }
            return count;
        }
    }
}