Java for LeetCode 057 Insert Interval

標籤：

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].

解題思路一：

參考Java for LeetCode 056 Merge Intervals思路一，去掉最外層迴圈即可，JAVA實現如下：

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {        if (newInterval == null)            return intervals;            int startIndex = 0, endIndex = 0;            for (int j = 0; j < intervals.size(); j++) {                if (newInterval.start > intervals.get(j).end) {                    startIndex += 2;                    endIndex += 2;                    continue;                }                if (newInterval.end < intervals.get(j).start)                    break;                if (newInterval.start >= intervals.get(j).start)                    startIndex++;                if (newInterval.end > intervals.get(j).end) {                    endIndex += 2;                    continue;                }                if (newInterval.end >= intervals.get(j).start)                    endIndex++;                break;            }            if(startIndex==endIndex&&startIndex%2==0)                intervals.add(startIndex/2,new Interval(newInterval.start,newInterval.end));            else if(startIndex%2==0&&endIndex%2==0){                intervals.get(startIndex/2).start=newInterval.start;                intervals.get(startIndex/2).end=newInterval.end;                for(int k=1;k<endIndex/2-startIndex/2;k++)                intervals.remove(startIndex/2+1);            }            else if(startIndex%2==0&&endIndex%2!=0){                intervals.get(startIndex/2).start=newInterval.start;                intervals.get(startIndex/2).end=intervals.get(endIndex/2).end;                for(int k=1;k<=endIndex/2-startIndex/2;k++)                    intervals.remove(startIndex/2+1);            }            else if(startIndex%2!=0&&endIndex%2==0){                intervals.get(startIndex/2).end=newInterval.end;                for(int k=1;k<endIndex/2-startIndex/2;k++)                    intervals.remove(startIndex/2+1);            }            else if(startIndex%2!=0&&endIndex%2!=0){                intervals.get(startIndex/2).end=intervals.get(endIndex/2).end;                for(int k=1;k<=endIndex/2-startIndex/2;k++)                    intervals.remove(startIndex/2+1);            }        return intervals;    }

 解題思路二：

參考Java for LeetCode 056 Merge Intervals思路二，添加後重新排序即可，JAVA實現如下：

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {        if (intervals == null)         return intervals;        List<Interval> list = new ArrayList<Interval>();        intervals.add(newInterval);        Comparator<Interval> comparator = new Comparator<Interval>() {            @Override            public int compare(Interval o1, Interval o2) {                if (o1.start == o2.start)                    return o1.end - o2.end;                return o1.start - o2.start;            }        };        Collections.sort(intervals, comparator);        for (Interval interval : intervals) {            if (list.size() == 0 || list.get(list.size() - 1).end < interval.start)                 list.add(new Interval(interval.start, interval.end));            else                list.get(list.size() - 1).end = Math.max(interval.end, list.get(list.size() - 1).end);        }        return list;    }

 

Java for LeetCode 057 Insert Interval