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]

.

/**
* 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) {
if (newInterval == null || intervals == null) {
return intervals;
}

List<Interval> results = new ArrayList<Interval>();
int insertPos = 0;

for (Interval interval : intervals) {
if (interval.end < newInterval.start) { //判断能不能插到后面，可以的话pos++
results.add(interval);
insertPos++;
} else if (interval.start > newInterval.end) {//判断能不能插到前面，pos不变
results.add(interval);
} else {
newInterval.start = Math.min(interval.start, newInterval.start); //都不行的话进行merge
newInterval.end = Math.max(interval.end, newInterval.end);
}
}

results.add(insertPos, newInterval);

return results;
}

}