Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
题意:
给定一些区间,将重叠部分合并。
思路:
将原interval集合里,给定区间按照start开头值的从小到大排序
建一个新的interval集合
遍历原interval集合的每个区间,
若cur.start > pre.end 则说明没有重叠,扔到新的interval集合去。
否则,有重叠,则更新之前区间.end的长度
代码:
class Solution {
public List<Interval> merge(List<Interval> intervals) {
Collections.sort(intervals, (o1, o2)-> o1.start - o2.start); List<Interval> result = new ArrayList<>(); Interval pre = null;
for(Interval cur : intervals){
if(pre == null || cur.start > pre.end){
result.add(cur);
pre = cur;
}else{
pre.end = Math.max(pre.end, cur.end);
}
}
return result;
}
}