[leetcode]Insert Interval

这道题。。。。简直了。。。

新博文地址：[leetcode]Insert Interval 

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].

 这道题不难，就是烦，考虑了各种cover的组合情况，代码繁杂，看着就不对劲，网上搜了一下，找到了这样一个犀利的解法，抽象、简洁、易懂，真经典

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
    	List<Interval> list = new LinkedList<Interval>();
    	if(intervals == null || intervals.isEmpty()){
    		list.add(newInterval);
    		return list;
    	}
    	if(newInterval == null) return intervals;
    	int newStart = newInterval.start,newEnd = newInterval.end;
    	int i = 0;
    	for(; i < intervals.size() && intervals.get(i).end < newInterval.start; i++){
    		list.add(intervals.get(i));
    	}
    	if(i < intervals.size()){
    		newStart = Math.min(intervals.get(i).start,newStart);
    	}
    	for(; i < intervals.size() && newEnd >= intervals.get(i).start; i++){
    		newEnd = Math.max(newEnd, intervals.get(i).end);
    	}
    	list.add(new Interval(newStart, newEnd));
    	for(;i < intervals.size(); i++){
    		list.add(intervals.get(i));
    	}
    	return list;
    }