算法描述：

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.

解題思路：排序，比較，注意邊界值，比較函數的編寫。比較函數必須是靜態函數。

vector<Interval> merge(vector<Interval>& intervals) {if(intervals.size() <=1) return intervals; vector<Interval> results;sort(intervals.begin(),intervals.end(),compare);int start = intervals[0].start;int end = intervals[0].end;for(int i = 0; i <intervals.size(); i++){if(intervals[i].start <= end){end = max(end,intervals[i].end);}else{Interval temp(start, end);results.push_back(temp);start =intervals[i].start;end= intervals[i].end;}}Interval temp(start, end);results.push_back(temp);return results;}static bool compare(Interval a, Interval b){if(a.start < b.start) return true;else return false;}

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轉載于:https://www.cnblogs.com/nobodywang/p/10337131.html

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