leetcode 57:Insert Interval

题目：

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].

思路：

这题思路和Merge Intervals比较像，由于已经排序，所以不需进行排序，然后插入集合，进行合并。

题目不难，但是注意边界条件即可。

时间复杂度：O(n)

实现如下：

class Solution {
public:
	vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
		int size = intervals.size();
		vector<Interval> result;
		if (size == 0)
		{
			result.push_back(newInterval);
			return result;
		}
		int i = 0,low,high;
		while (i<size && intervals[i].end < newInterval.start)
		{
			result.push_back(intervals[i]);
			++i;
		}
		if (i<size && newInterval.end < intervals[i].start)
			result.push_back(newInterval);
		else
		{
			low = i < size ? min(intervals[i].start, newInterval.start) : newInterval.start;
			while (i < size && newInterval.end >= intervals[i].start) ++i;
			high = i>0 ? max(intervals[i-1].end, newInterval.end) : newInterval.end;
			Interval temp(low, high);
			result.push_back(temp);
		}
		while (i < size)
		{
			result.push_back(intervals[i]);
			++i;
		}
		return result;
	}
};

另外一种思路是：

先将newInterval插入到intervals，然后再对intervals进行合并。

运用到Merge Intervals的解法。

实现如下：

class Solution {
public:
	static bool comp(const Interval a, const Interval b)
	{
		return a.start < b.start ? 1 : 0;
	}
	vector<Interval> insert(vector<Interval>& intervals,Interval newInterval) {
		intervals.push_back(newInterval);
		vector<Interval> result;
		int size = intervals.size();
		if (size < 2) return intervals;
		sort(intervals.begin(), intervals.end(), comp);
		for (int i = 0; i<size; ++i)
		{
			Interval temp(intervals[i].start, intervals[i].end);
			int high = intervals[i].end;
			while (i < size - 1 && high >= intervals[i + 1].start)
			{
				if (intervals[i + 1].end > high) high = intervals[i + 1].end;
				++i;
			}
			temp.end = high;
			result.push_back(temp);
		}
		return result;
	}
};