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Insert Interval leetcode java
题目:
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]
.
题解:
这道题不仅要insert newInterval同时还要保证能够merge。那么就分情况讨论。
遍历每一个已给出的interval,
当当前的interval的end小于newInterval的start时,说明新的区间在当前遍历到的区间的后面,并且没有重叠,所以res添加当前的interval;
当当前的interval的start大于newInterval的end时,说明新的区间比当前遍历到的区间要前面,并且也没有重叠,所以把newInterval添加到res里,并更新newInterval为当前的interval;
当当前的interval与newInterval有重叠时,merge interval并更新新的newInterval为merge后的。
代码如下:
2 ArrayList<Interval> res = new ArrayList<Interval>();
3
4 for(Interval each: intervals){
5 if(each.end < newInterval.start){
6 res.add(each);
7 }else if(each.start > newInterval.end){
8 res.add(newInterval);
9 newInterval = each;
10 }else if(each.end >= newInterval.start || each.start <= newInterval.end){
11 int nstart = Math.min(each.start, newInterval.start);
12 int nend = Math.max(newInterval.end, each.end);
13 newInterval = new Interval(nstart, nend);
14 }
15 }
16
17 res.add(newInterval);
18
19 return res;
20 }
Reference://http://www.programcreek.com/2012/12/leetcode-insert-interval/