LeetCode题目 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].

解析：对于这个问题无非就三种情况：1、被给出的间隔的终端比要插入的起始端小，则直接将其插入。2、被给出的间隔的起始端比要插入的终端大，将两者都插入。3、介入上述两种情况之间 即存在交叉。则考虑起始和终端时应取最小的起始端和最大的终止端即可。

代码：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> result;
    vector<Interval>::iterator it;
	bool flag=true;
	for (it=intervals.begin();it!=intervals.end();it++)
	{
		if (it->end<newInterval.start)
		{
			result.push_back(*it);
			continue;
		}
		if (it->start>newInterval.end)
		{
			if (flag)
			{
				result.push_back(newInterval);
				flag=false;
			}
			
			result.push_back(*it);
			continue;
		}
		newInterval.start=it->start<newInterval.start?it->start:newInterval.start;
		newInterval.end=it->end>newInterval.end?it->end:newInterval.end;

	}
	if (flag)
	{
		result.push_back(newInterval);
	}
	return result;
        
    }
};