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;
    }
}