56 - Merge Intervals

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].

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思路分析：

To check the intersections between interval [a,b] and [c,d],  there are four cases (equal not shown in the figures):

    a____b

c____d

a____b

     c____d

a_______b

    c___d

   a___b

c_______d

But we can simplify these into 2 cases when check the smaller (smaller start point) interval with the bigger interval.

For the problem, the idea is simple. First sort the vector according to the start value. Second, scan every interval, if it can be merged to the previous one, then merge them, else push it into the result vector.

Note here:
The use of std::sort to sort the vector, need to define a compare function, which need to be static. (static bool myfunc() ) The sort command should be like this:  std::sort(intervals.begin,intervals.end, Solution::myfunc); otherwise, it won't work properly.

/*
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].
*/

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <algorithm>



using namespace std;

struct Interval 
{
	int start;
	int end;
	Interval() : start(0), end(0) {}
	Interval(int s, int e) : start(s), end(e) {}
	
};

class Solution_056_MergeIntervals
{
public:
	static bool myfunc(const Interval &a, const Interval &b)
	{
		return (a.start < b.start);
	}

	vector<Interval> merge(vector<Interval> &intervals)
	{
		vector<Interval> res;

		if (intervals.size() == 0)
		{
			return res;
		}

		sort(intervals.begin(), intervals.end(), myfunc);
		res.push_back(intervals[0]);

		for (int i = 1; i < intervals.size(); i++)
		{
			/*
				a_____b
				    c____d
				或者
				a____b
				  c_d
			*/
			if (res.back().end >= intervals[i].start)
			{
				res.back().end = max(res.back().end, intervals[i].end);
			}

			/*
				a____b
						c____d
			
			*/
			else
			{
				res.push_back(intervals[i]);
			}
		}

		return res;

	}

};