452. Minimum Number of Arrows to Burst Balloons

最新推荐文章于 2023-11-18 01:25:37 发布

原创最新推荐文章于 2023-11-18 01:25:37 发布 · 640 阅读

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探讨如何通过最少次数的射箭来爆破所有水平分布的气球。文章介绍了一种排序方法，并提供了两种解决方案：一是使用最小堆处理结束坐标；二是简化为单一变量跟踪最小结束坐标。

原题网址：https://leetcode.com/problems/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).

思路：与会议室、排课表问题类似，先对开始坐标进行排序。当遇到一个结束坐标时，该气球必须被戳爆，顺带戳爆其他的气球。

方法一：先对开始坐标排序，再用最小堆处理结束坐标。

public class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                return a[0] - b[0];
            }
        });
        
        PriorityQueue<int[]> minHeap = new PriorityQueue<>(new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                return a[1] - b[1];
            }
        });
        int shots = 0;
        for(int i = 0; i < points.length; i++) {
            if (!minHeap.isEmpty() && points[i][0] > minHeap.peek()[1]) {
                shots++;
                minHeap.clear();
            }
            minHeap.offer(points[i]);
        }
        if (!minHeap.isEmpty()) {
            shots++;
        }
        return shots;
    }
}

方法二：最小堆可以简化为一个变量。

public class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                return a[0] - b[0];
            }
        });

        int shots = 0;
        boolean hasMin = false;
        int minValue = Integer.MIN_VALUE;
        for(int i = 0; i < points.length; i++) {
            if (!hasMin) {
                minValue = points[i][1];
                hasMin = true;
            } else if (points[i][0] > minValue) {
                shots++;
                minValue = points[i][1];
            } else {
                minValue = Math.min(minValue, points[i][1]);
            }
        }
        if (hasMin) {
            shots++;
        }
        return shots;
    }
}