leetcode 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].

区间合并，这个和上一道题的做法一样，这里我们有旋转插入排序，而是直接添加然后排序，最后合并。

代码如下：

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;

/*class Interval
{
int start;
int end;
Interval() { start = 0; end = 0; }
Interval(int s, int e) { start = s; end = e; }
}*/

public class Solution
{
public List<Interval> insert(List<Interval> intervals, Interval newInterval)
{
intervals.add(newInterval);
if(intervals==null || intervals.size()<=1)
return intervals;

List<Interval> res=new ArrayList<Interval>();
intervals.sort(new Comparator<Interval>() {
@Override
public int compare(Interval o1, Interval o2) {
return o1.start-o2.start;
}
});

Interval pre=intervals.get(0);
for(int i=1;i<intervals.size();i++)
{
Interval now=intervals.get(i);
if(pre.end>=now.start)
pre=new Interval(pre.start, Math.max(pre.end, now.end));
else
{
res.add(pre);
pre=now;
}
}
res.add(pre);
return res;
}
}

和上一道题一样，先插入，然后排序，最后合并即可

代码如下：

#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) {}
};
*/

bool cmp(Interval a, Interval b)
{
if (a.start < b.start)
return true;
else
return false;
}

class Solution
{
public:
vector<Interval> insert(vector<Interval>& in, Interval newInterval)
{
in.push_back(newInterval);
if (in.size() <= 1)
return in;

vector<Interval> res;
sort(in.begin(),in.end(),cmp);

Interval pre = in[0];
for (int i = 1; i < in.size(); i++)
{
Interval now = in[i];
if (pre.end < now.start)
{
res.push_back(pre);
pre = now;
}
else
pre = Interval(pre.start,max(pre.end,now.end));
}
res.push_back(pre);
return res;
}
};