59. Employee Free Time（贪心）

最新推荐文章于 2020-09-16 03:44:16 发布

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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。输入为每个员工的工作时间区间列表，输出为所有员工共同的正长度空闲时间区间列表。通过合并所有员工的时间区间并排序，再遍历这些区间来确定空闲时间。

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1.题目描述

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.

2. 代码

分析：将所有的区间放到一起，按开始时间从小到大排序（如果开始时间相同则根据结束时间的先后，思考为什么？），这样处理后，进行一次遍历.

如果两个区间之间没有空闲时间而存在交集，那么将这两个交集合二为一（取两者的最小开始时间和最大结束时间）

如果存在空闲区间，那么将该区间放入结果中！

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    vector<Interval> employeeFreeTime(vector<vector<Interval>>& schedule) {
        vector<Interval> all;

        for(vector<Interval> tmp : schedule) {
            for(Interval s : tmp) {
                all.push_back(s);
            }
        }

        sort(all.begin(), all.end(), comp);

        for(int i = 0; i<all.size(); ++i)
            cout << "[" << all[i].start << "," << all[i].end << "]" << endl; 
        vector<Interval> res;
        for(int i = 0; i < all.size()-1; ++i) {
            if(all[i].end < all[i+1].start) {
                cout << all[i].end << ", " << all[i+1].start << endl;
                Interval interval(all[i].end, all[i+1].start);
                res.push_back(interval);
            } else {
                all[i+1].start = min(all[i].start, all[i+1].start);
                all[i+1].end = max(all[i].end, all[i+1].end);
            }   
        }
        return res;

    }

    static bool comp(const Interval& s1, const Interval& s2) {
        if(s1.start < s2.start) return true;
        else if(s1.start == s2.start) {
            if(s1.end <= s2.end) return true;
            else return false; 
        }
        else return false;  
    }
};

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