线段区间问题

**56. 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].
代码如下：

class Solution {
public:
    static bool compare(Interval a,Interval b)
    {
        return a.start<b.start;
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        vector<Interval> rec;
        if(intervals.size()<=0)
          return rec;
        sort(intervals.begin(),intervals.end(),compare);
        rec.push_back(intervals[0]);
        for(int i=1;i<intervals.size();i++)
        {
            if(intervals[i].start<=rec[rec.size()-1].end)
               rec[rec.size()-1].end = max(rec[rec.size()-1].end,intervals[i].end);
            else
               rec.push_back(intervals[i]);
        }
        return rec;
    }
};

**57. Insert Interval **
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
代码如下：

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        vector<Interval> rec;
        if(intervals.empty())
        {
            rec.push_back(newInterval);
            return rec;
        }
        for(int i=0;i<intervals.size();i++)
        {
            if(newInterval.end<intervals[i].start)
            {
                rec.push_back(newInterval);
                while(i<intervals.size())
                  rec.push_back(intervals[i++]);
                return rec;
            }
            else if(newInterval.start>intervals[i].end)
                rec.push_back(intervals[i]);
            else
            {
                newInterval.start = min(newInterval.start,intervals[i].start);
                newInterval.end = max(newInterval.end,intervals[i].end);
            }
        }
        rec.push_back(newInterval);
        return rec;
    }
};