LeetCode OJ算法题（五十六）：Insert Interval

题目：

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].

解法：

题目给出的区间已经按start排好序了，那么我们只需要处理插入的区间是在最前面，最后面，两个区间之间，以及会出现融合的四种情况。

遍历所有区间，比较newInterval的start与当前元素的end，以确定插入的起始位置，比较newtiterval的end与当前元素的start，以确定融合区间所包含的范围

public static List<Interval> insert(List<Interval> intervals, Interval newInterval) {        List<Interval> ret = new ArrayList<Interval>();        int i=0;        while(i<intervals.size() && intervals.get(i).end<newInterval.start)        ret.add(intervals.get(i++));        if(i == intervals.size()){        ret.add(newInterval);        return ret;        }        Interval tmp = intervals.get(i);        if(tmp.start > newInterval.end){        ret.add(newInterval);        while(i<intervals.size())            ret.add(intervals.get(i++));        return ret;        }        tmp.start = tmp.start<newInterval.start?tmp.start:newInterval.start;        while(i<intervals.size() && intervals.get(i).start<=newInterval.end){        i++;        }        tmp.end = newInterval.end>intervals.get(i-1).end?newInterval.end:intervals.get(i-1).end;        ret.add(tmp);        while(i<intervals.size())        ret.add(intervals.get(i++));        return ret;    }