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].

思路：

先对数据进行排序，然后逐个插入。注意在使用sort的时候，最开始出现<unresolved overloaded function type>错误，后来把myCmp改成static的以后就解决了。

代码：

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

 第二遍：

    static bool myCmp(const Interval &a, const Interval &b){
        if(a.start == b.start)
            return a.end < b.end;
        return a.start < b.start;
    }
    vector<Interval> merge(vector<Interval> &intervals) {
        vector<Interval> result;
        if(!intervals.empty()){
            sort(intervals.begin(), intervals.end(), myCmp);
            int start = intervals[0].start, end = intervals[0].end;
            for(int i = 1; i < intervals.size(); i++){
                if(intervals[i].start <= end)
                    end = end > intervals[i].end ? end : intervals[i].end;
                else{
                    result.push_back(Interval(start, end));
                    start = intervals[i].start;
                    end = intervals[i].end;
                }
            }
            result.push_back(Interval(start, end));
        }
        return result;
    }