LeetCode-56-Merge Intervals

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.

解题思路：排序，比较，注意边界值，比较函数的编写。比较函数必须是静态函数。

    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;
    }