LeetCode 96 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].

分析：

end < newStart 的区间都在前面；

start > newEnd 的区间都在后面；

交叉的区间扩充。

/**
* 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 ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {

if(intervals == null || newInterval == null)
return intervals;

Interval temp = newInterval;
ArrayList<Interval> result = new ArrayList<Interval>();

for(Interval item : intervals){
//在前
if(item.end < temp.start)
result.add(item);
//在后
else if(item.start > temp.end){
result.add(temp);
temp = item;
//交叉
}else{
temp.start = Math.min(item.start, temp.start);
temp.end = Math.max(item.end, temp.end);
}
}
result.add(temp);

return result;
}
}