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].

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> result = new ArrayList<Interval>();
intervals.add(newInterval);
intervals.sort((a,b)->(a.start - b.start));
for(Interval currentInterval:intervals ){
if(result.size() == 0){
result.add(currentInterval);
}
else{
Interval lastInterval = result.get(result.size()-1);
if(currentInterval.start > lastInterval.end) result.add(currentInterval);
else if(currentInterval.end > lastInterval.end){
lastInterval.end = currentInterval.end;
}
}
}
return result;
}

只加了intervals.add(newInterval);

没有做优化

实际上 由于原来是有序的 不需要插入重排

应该logn找到对应位置 局部处理即可