452. Minimum Number of Arrows to Burst Balloons Java解法

本文链接：https://blog.youkuaiyun.com/sysu_jiang/article/details/72295218

探讨如何计算使所有二维平面上分布的圆形气球爆破所需的最少垂直箭矢射击次数。通过一种特定的排序及遍历算法，确定能够同时击中多个气球的最佳射击位置。

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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).

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) {
                return a[1] - b[1];
            }
        });

        long lastEnd = Long.MIN_VALUE;
        int minArrows = 0;
        for (int i = 0; i < points.length; i++) {
            if (lastEnd < points[i][0]) {
                lastEnd = points[i][1];
                minArrows++;
            }
        }
        return minArrows;
    }