标签：leetcode sid == 题解 else opera font 图片 ase
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.
合并重合的闭区间
1 bool operator <(const Interval& a, const Interval& b) { 2 if (a.start == b.start) 3 return a.end > b.end; 4 return a.start < b.start; 5 } 6 7 class Solution { 8 public: 9 vector<Interval> merge(vector<Interval>& intervals) { 10 if (intervals.size()<=1)return intervals; 11 set<Interval>q; 12 while (!intervals.empty()) { 13 Interval now = intervals.back(); 14 intervals.pop_back(); 15 q.insert(now); 16 } 17 auto pre = q.begin(); 18 int mm = (*pre).end; 19 for (auto p = q.begin(); p != q.end(); ) { 20 if (p != q.begin() && (*p).start <= (*pre).end) { 21 mm = max((*p).end, mm); 22 auto oldp = p; 23 p++; 24 q.erase(oldp); 25 } 26 else if (p != q.begin()) { 27 int star = (*pre).start; 28 q.erase(pre); 29 q.insert(Interval(star, mm)); 30 if ((*p).start <= mm) 31 pre = q.find(Interval(star, mm)); 32 else { 33 pre = p; 34 p++; 35 } 36 mm = (*pre).end; 37 } 38 else p++; 39 } 40 if ((*pre).end < mm) { 41 q.insert(Interval((*pre).start, mm)); 42 q.erase(pre); 43 } 44 return vector<Interval>(q.begin(),q.end()); 45 } 46 };
19.2.8 [LeetCode 56] Merge Intervals
标签：leetcode sid == 题解 else opera font 图片 ase
原文地址：https://www.cnblogs.com/yalphait/p/10356600.html
