LWC 66: 759. Employee Free Time

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最新推荐文章于 2020-09-16 03:44:16 发布

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分类专栏：算法竞赛算法集中营

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本文介绍了一种算法，用于计算一组员工的共同空闲时间段。通过将所有员工的工作时间区间合并，并找出这些区间之间的空闲时段。算法首先按区间的开始时间进行排序，然后寻找不重叠的时间段作为员工们的共同空闲时间。

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LWC 66: 759. Employee Free Time

传送门：759. Employee Free Time

Problem:

We are given a list schedule of employees, which represents the working time for each employee.

Each employee has a list of non-overlapping Intervals, and these intervals are in sorted order.

Return the list of finite intervals representing common, positive-length free time for all employees, also in sorted order.

Example 1:

Input: schedule = [[[1,2],[5,6]],[[1,3]],[[4,10]]]
Output: [[3,4]]
Explanation:
There are a total of three employees, and all common
free time intervals would be [-inf, 1], [3, 4], [10, inf].
We discard any intervals that contain inf as they aren’t finite.

Example 2:

Input: schedule = [[[1,3],[6,7]],[[2,4]],[[2,5],[9,12]]]
Output: [[5,6],[7,9]]
(Even though we are representing Intervals in the form [x, y], the objects inside are Intervals, not lists or arrays. For example, schedule[0][0].start = 1, schedule[0][0].end = 2, and schedule[0][0][0] is not defined.)

Also, we wouldn’t include intervals like [5, 5] in our answer, as they have zero length.

Note:

schedule and schedule[i] are lists with lengths in range [1, 50].
0 <= schedule[i].start < schedule[i].end <= 10^8.

思路：
大致题意是求所有区间的并集的补集。先根据区间的start排序，假设当前区间为now，那么如果新来的区间start在now区间表示的范围内，说明两个区间存在交集，更新右边界，直到不存在交集时，能够求得第一个补集，依此类推。

Java版本：

    class P implements Comparable<P>{
        int start;
        int end;

        P(int start, int end){
            this.start = start;
            this.end   = end;
        }

        @Override
        public int compareTo(P o) {
            return this.start == o.start ? this.end - o.end : this.start - o.start;
        }

        @Override
        public String toString() {
            return "[" + start + ", " + end + "]";
        }
    }

    public List<Interval> employeeFreeTime(List<List<Interval>> avails) {
        List<P> unSorted = new ArrayList<>();
        for (List<Interval> avail : avails) {
            for (Interval inter : avail) {
                unSorted.add(new P(inter.start, inter.end));
            }
        }
        Collections.sort(unSorted);
        List<Interval> ans = new ArrayList<>();
        int n = unSorted.size();
        for (int i = 0; i < n; ++i) {
            P now = unSorted.get(i);
            int maxRight = now.end;
            i ++;
            while (i < n && unSorted.get(i).start <= maxRight) {
                maxRight = Math.max(maxRight, unSorted.get(i).end);
                i ++;
            }
            if (i < n) ans.add(new Interval(maxRight, unSorted.get(i).start));
            i --;
        }
        return ans;
    }

Python版本：

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
    def employeeFreeTime(self, avails):
        """
        :type schedule: List[List[Interval]]
        :rtype: List[Interval]
        """
        from itertools import chain
        avails = list(sorted(chain(*avails), key = lambda Interval : Interval.start))
        ans = list()
        n = len(avails)
        i = 0
        while (i < n):
            now = avails[i]
            maxRight = now.end
            j = i + 1
            while (j < n and avails[j].start <= maxRight):
                maxRight = max(maxRight, avails[j].end)
                j += 1
            i = j - 1
            if j < n: ans.append(Interval(maxRight, avails[j].start))
            i += 1
        return ans