leetcode452. Minimum Number of Arrows to Burst Balloons

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).

解法一

按照end进行升序排序，判断下一个start是否大于该end，如果大于，ans加一。

public class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points == null || points.length == 0) {
            return 0;
        }
        if (points[0] == null || points[0].length == 0) {
            return 0;
        }

        Arrays.sort(points, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return o1[1] - o2[1];
            }
        });
        int terminal = points[0][1];
        int ans = 1;
        for (int i = 1; i < points.length; i++) {
            if (points[i][0] > terminal) {
                ans++;
                terminal = points[i][1];
            }
        }
        return ans;
    }
}

解法二

按照start升序排序，判断下一个区间的start是否大于该区间的terminal，如果大于，terminal加大，ans加1；如果不大于，是否下一个区间的terminal是否小于该区间的terminal，即是否嵌套，如果小于，terminal变小，变为下一个区间的terminal。

public class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points == null || points.length == 0) {
            return 0;
        }
        if (points[0] == null || points[0].length == 0) {
            return 0;
        }

        Arrays.sort(points, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return o1[0] - o2[0];
            }
        });
        int terminal = points[0][1];
        int ans = 1;
        for (int i = 1; i < points.length; i++) {
            if (points[i][0] > terminal) {
                ans++;
                terminal = points[i][1];
            } else if (points[i][1] < terminal) {
                terminal = points[i][1];
            }
        }
        return ans;
    }
}