Insert Interval

Insert Interval

问题：

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.

思路：

　　插入位置+merge

我的代码：

public class Solution {

    public List<Interval> merge(List<Interval> intervals) {

        if(intervals == null || intervals.size() == 0)  return intervals;

        Collections.sort(intervals, new IntervalComparator());

        for(int i = 0; i < intervals.size() - 1; i++)

        {

            Interval one = intervals.get(i);

            Interval two = intervals.get(i+1);

            if(one.end >= two.start)

            {

                one.end = Math.max(one.end, two.end);

                intervals.remove(i+1);

                i--;

            }

        }

        return intervals;

    }

    private class IntervalComparator implements Comparator<Interval>

    {

        public int compare(Interval a, Interval b)

        {

            return a.start - b.start;

        }

    }

}

View Code

他人代码：

public class Solution {

    public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {

        if (newInterval == null || intervals == null) {

            return intervals;

        }



        ArrayList<Interval> results = new ArrayList<Interval>();

        int insertPos = 0;



        for (Interval interval : intervals) {

            if (interval.end < newInterval.start) {

                results.add(interval);

                insertPos++;

            } else if (interval.start > newInterval.end) {

                results.add(interval);

            } else {

                newInterval.start = Math.min(interval.start, newInterval.start);

                newInterval.end = Math.max(interval.end, newInterval.end);

            }

        }



        results.add(insertPos, newInterval);



        return results;

    }

}

View Code

学习之处：

• 我的代码需要进行两遍的扫描，别人代码需要一遍扫描，我的空间复杂度是O(1)，别人的代码空间复杂度是O(n)