The idea to solve this problem is to first sort the intervals according to their start field and then scan the intervals from head to tail and merge those that overlap.
For sorting the intervals according to their start field, we define a comparison function as follows.
1 static bool comp(Interval interval1, Interval interval2) { 2 return interval1.start < interval2.start; 3 }
Then all the intervals are sorted in the ascending order of start. Now we define a current intervalcur and initialize it to be intervals[0]. Then we scan from intervals[1] to intervals[n - 1]. If it overlaps with cur, merge them; otherwise, add cur to res, update cur to be intervals[i]and move on with the merging process.
There are two required subroutines in the above process: isOverlap to tell whether two intervals overlap and mergeTwo to merge two overlapping intervals.
For isOverlap, since the intervals are sorted in ascending order of start, we simply need to guarantee that end of the left (with smaller start) interval is not less than start of the right (with larger start) interval.
For mergeTwo, just take the minimum of start and maximum of end of the two overlapping intervals and return a new interval with these two values.
The complete code is as follows, which should be self-explanatory.
1 class Solution { 2 public: 3 vector<Interval> merge(vector<Interval>& intervals) { 4 vector<Interval> res; 5 if (intervals.empty()) return res; 6 sort(intervals.begin(), intervals.end(), comp); 7 Interval cur(intervals[0].start, intervals[0].end); 8 for (int i = 1, n = intervals.size(); i < n; i++) { 9 if (isOverlap(cur, intervals[i])) 10 cur = mergeTwo(cur, intervals[i]); 11 else { 12 res.push_back(cur); 13 cur = intervals[i]; 14 } 15 } 16 res.push_back(cur); 17 return res; 18 } 19 private: 20 static bool comp(Interval interval1, Interval interval2) { 21 return interval1.start < interval2.start; 22 } 23 bool isOverlap(Interval interval1, Interval interval2) { 24 return interval1.end >= interval2.start; 25 } 26 Interval mergeTwo(Interval interval1, Interval interval2) { 27 int start = min(interval1.start, interval2.start); 28 int end = max(interval1.end, interval2.end); 29 return Interval(start, end); 30 } 31 };
