19.2.8 [LeetCode 56] Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

题意

合并重合的闭区间

题解

bool operator <(const Interval& a, const Interval& b) {
    if (a.start == b.start)
        return a.end > b.end;
    return a.start < b.start;
}

class Solution {
public:
    vector<Interval> merge(vector<Interval>& intervals) {
        if (intervals.size()<=1)return intervals;
        set<Interval>q;
        while (!intervals.empty()) {
            Interval now = intervals.back();
            intervals.pop_back();
            q.insert(now);
        }
        auto pre = q.begin();
        int mm = (*pre).end;
        for (auto p = q.begin(); p != q.end(); ) {
            if (p != q.begin() && (*p).start <= (*pre).end) {
                mm = max((*p).end, mm);
                auto oldp = p;
                p++;
                q.erase(oldp);
            }
            else if (p != q.begin()) {
                int star = (*pre).start;
                q.erase(pre);
                q.insert(Interval(star, mm));
                if ((*p).start <= mm) 
                    pre = q.find(Interval(star, mm));
                else {
                    pre = p;
                    p++;
                }
                mm = (*pre).end;
            }
            else p++;
        }
        if ((*pre).end < mm) {
            q.insert(Interval((*pre).start, mm));
            q.erase(pre);
        }
        return vector<Interval>(q.begin(),q.end());
    }
};