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]

.

思路：在list中删除在start和end之间的元素，如果start，或end在某一元素的start和end之间，删除该元素，调整插入元素的start或end值。最后将所需插入的元素插入合适的位置。思想很简单，细节上调了很久，很坑的一题。

/**
* Definition for an interval.
* public 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) {
int n=intervals.size();
int index=0;

for(int i=0;i<n;i++){
Interval temp=null;

temp=intervals.get(i);
if(newInterval.start>=temp.start&&newInterval.end<=temp.end){
return intervals;
}else
if(newInterval.start>=temp.start&&newInterval.start<=temp.end){
newInterval.start=temp.start;
intervals.remove(i);
index=i;
n--;
i--;

}else
if(temp.start>newInterval.start&&temp.end<newInterval.end){
intervals.remove(i);
index=i;
n--;
i--;

}else
if(newInterval.end>=temp.start&&newInterval.end<=temp.end){
newInterval.end=temp.end;
intervals.remove(i);
index=i;
n--;
i--;

}

}
if(n==0){
intervals.add(newInterval);
return intervals;
}

if(newInterval.end<intervals.get(0).start){
intervals.add(0,newInterval);
return    intervals;
}else if(newInterval.start>intervals.get(n-1).end){
intervals.add(newInterval);
return    intervals;
}

for(int j=0;j<n-1;j++){
Interval temp1=intervals.get(j);
Interval temp2=intervals.get(j+1);
if(temp1.end<newInterval.start&&temp2.start>newInterval.end){
intervals.add(j+1,newInterval);
}
}
return    intervals;
}
}