Leetcode56.Merge_Intervals

先将Intervals按start的值排列，然后从头遍历数组，将后方start小于前方end的interval相合并。如果后边元素的start大于前边元素的end，前方元素向后移一位。
时间复杂度:O(NlogN)
C++代码：

class Solution {
	static bool compare(Interval i1, Interval i2)
	{
		return (i1.start < i2.start) ? true : false;
	}
public:
	vector<Interval> merge(vector<Interval>& intervals) {
		if (intervals.empty())
			return {};
		if (intervals.size() == 1)
			return intervals;
		sort(intervals.begin(), intervals.end(), compare);
		auto it = intervals.begin();
		for (int i = 0; i < intervals.size(); i++)
		{
			if (it == intervals.begin() + i)
				continue;
			if (it->end >= intervals[i].start)
			{
				if (it->end < intervals[i].end)
					it->end = intervals[i].end;
				intervals.erase(intervals.begin() + i);
				i--;
			}
			else
				it++;
		}
		return intervals;
	}
};