leetcode 57. Insert Interval-优快云博客

本文介绍了一种插入并合并区间的高效算法。该算法适用于已排序的非重叠区间集合，通过一次循环处理新插入的区间并与现有区间进行合并。文章提供了一个具体的实现示例，展示了如何在保持时间效率的同时正确地合并区间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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].

一个插入合并区间的题，虽然是hard，但相对简单。

已有区间是有序的,找到要插入的位置即可，注意可能有多个区间需要合并

本来的思路是找到原有区间需要与新区间合并的合并到一起形成一个大的，然后把原来的那个区间remove掉，但remove操作是一个O(n)的操作，虽然ac了但出奇的慢。

改成一次循环，取出旧区间、合并、add到新的list后效率提升不少

class Solution {
public static List<Interval> insert(List<Interval> intervals, Interval newInterval) {

List<Interval> list = new ArrayList<Interval>();
boolean found=false,end=false;
for(int i=0;i<intervals.size();i++){
if(end){
list.add(intervals.get(i));
}
else if(found && !(intervals.get(i).start >=newInterval.start &&intervals.get(i).start <=newInterval.end ||
intervals.get(i).end <=newInterval.end &&intervals.get(i).end >=newInterval.start||
intervals.get(i).end >=newInterval.end &&intervals.get(i).start <=newInterval.start)){
end=true;
list.add(newInterval);
list.add(intervals.get(i));
}
else if(intervals.get(i).start >=newInterval.start &&intervals.get(i).start <=newInterval.end ||
intervals.get(i).end <=newInterval.end &&intervals.get(i).end >=newInterval.start||
intervals.get(i).end >=newInterval.end &&intervals.get(i).start <=newInterval.start){

newInterval = merge(intervals.get(i),newInterval);

found=true;
}
else if(intervals.get(i).start > newInterval.end){
end=true;
list.add(newInterval);
list.add(intervals.get(i));
}
else
list.add(intervals.get(i));
}
if(!end)
list.add(newInterval);
return list;

}
public static Interval merge(Interval interval, Interval newInterval) {
Interval temp = new Interval();
temp.start = interval.start>=newInterval.start? newInterval.start:interval.start;
temp.end= interval.end<=newInterval.end? newInterval.end : interval.end;

return temp;

}
}