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]
.
解题思路一:
参考Java for LeetCode 056 Merge Intervals思路一,去掉最外层循环即可,JAVA实现如下:
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; }
解题思路二:
参考Java for LeetCode 056 Merge Intervals思路二,添加后重新排序即可,JAVA实现如下:
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; }
时间: 2024-08-09 23:52:37