Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

10/7/2015 update

Runtime error, quite strange.

I don’t figure out why yet.

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
struct {
    bool operator()(const pair<int, bool> a, const pair<int, bool> b){
        if(a.first < b.first){
            return true;
        }
        else if(a.first > b.first){
            return false;
        }
        else{
            if(a.second == true){
                return true;
            }
            else{
                return false;
            }
        }
    }
    
} cmp;
class Solution {
public:
    vector<Interval> merge(vector<Interval>& intervals) {
        vector<Interval> ans;
        vector<pair<int, bool> > points; //x, true means start point, false means end point
        for(auto interval : intervals){
            points.push_back(make_pair(interval.start, true));
            points.push_back(make_pair(interval.end, false));
        }
        sort(points.begin(), points.end(), cmp);
        int count = 0;
        int start = -1;
        for(auto it = points.begin(); it != points.end(); it++){
            auto point = *it;
            if(point.second == true){
                //left point
                if(start == -1){
                    start = point.first;
                }
                count++;
            }
            else{
                //right point
                count--;
                if(count == 0){
                    //end of a line segment
                    if(point.first > start){
                        ans.push_back(Interval(start, point.first));
                        
                    }
                    start = -1;
                }
            }
        }
        return ans;
    }
};

很有趣的题目。思路是，把给定的intervals数组按照start排序。然后遍历一下排序后的数组，比较当前元素的start是否落在了起始的(start, end)内，如果落在范围内的话，再把当前元素的end和起始的end作比较，如果当前元素end更大的话，则起始end=当前的end。达到扩展范围的目的。

需要注意，在for循环结束或，我们需要把最后一组start，end压入re中。

时间花在了排序上所以是o(nlogn)

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
struct{
    bool operator()(Interval a, Interval b){
        return a.start < b.start;
    }
}cmp;
class Solution {
public:
    vector<Interval> merge(vector<Interval> &intervals) {
        vector<Interval> re;
        if(intervals.empty()) return re;
        sort(intervals.begin(),intervals.end(), cmp);
        int start = intervals[0].start;
        int end = intervals[0].end;
        for(vector<Interval>::iterator it = intervals.begin(); it != intervals.end(); it++){
            //overlap, extend 'end'
            if((*it).start <= end){
                end = (*it).end > end? (*it).end: end;
            }
            //not overlap
            else{
                Interval tmp = Interval(start, end);
                re.push_back(tmp);
                start = (*it).start;
                end = (*it).end;
            }
        }
        Interval tmp = Interval(start, end);
        re.push_back(tmp);
        return re;
    }
};