Number of Airplanes in the Sky

最新推荐文章于 2020-09-13 00:10:14 发布

原创最新推荐文章于 2020-09-13 00:10:14 发布 · 458 阅读

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本文介绍了一种解决飞机起降时间重叠问题的算法。通过将起飞和降落时间转化为点，并利用比较器对这些时间点进行排序，可以计算出天空中最多同时有多少架飞机。该算法适用于飞行调度和航空交通管制等问题。

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线扫描问题

Given an interval list which are flying and landing time of the flight. How many airplanes are on the sky at most?

Notice

If landing and flying happens at the same time, we consider landing should happen at first.

Have you met this question in a real interview?

Yes

Example

For interval list

[
  (1,10),
  (2,3),
  (5,8),
  (4,7)
]

Return 3

/**
 * Definition of Interval:
 * public classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this.start = start;
 *         this.end = end;
 *     }
 * }
 */
class Point {
    int time;
    int flag;
    public Point(int time, int flag) {
        this.time = time;
        this.flag = flag;
    }
}

public class Solution {
    /**
     * @param airplanes: An interval array
     * @return: Count of airplanes are in the sky.
     */
    public int countOfAirplanes(List<Interval> airplanes) {
        // write your code here
        if (airplanes == null || airplanes.size() == 0) {
            return 0;
        }
        Comparator<Point> cmp = new Comparator<Point>() {
            public int compare(Point p1, Point p2) {
                if (p1.time == p2.time) {
                    return p1.flag - p2.flag;
                }
                return p1.time - p2.time;
            }
        };
        List<Point> list = new ArrayList<>();
        for (int i = 0; i < airplanes.size(); i++) {
            Interval val = airplanes.get(i);
            list.add(new Point(val.start, 1));
            list.add(new Point(val.end, 0));
        }
        Collections.sort(list, cmp);
        int count = 0, max = Integer.MIN_VALUE;
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i).flag == 1) {
                count++;
            } else {
                count--;
            }
            max = Math.max(max, count);
        }
        return max;
    }
}